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FIRST    LESSONS 


ALGEBRA, 


EMBRACING     THE     ELEMENTS 


SCIENCE. 


BY  CHARLES  DAVIES, 

ITJTHOR  OF  MENTAL  AND  PRACTICAL  ARITHMETIC,    ELEMENTS  OF  SURVKYIKG, 

BI.EMENTS   OF   DESCRIPTIVE   AND   ANALYTICAL   GEOMETRY,   ELEMENTS   OP 

DIFFERENTIAL   AND   INTEGRAL   CALCULUS,   AND   A  TREATISE   ON 

SHAPES,   SHADOWS   AND   PERSPECTIVE. 


PUBLISHED  BY 

A.  S.  BARNES  &   Co,  Hartford.— WILEY  &  PUTNAM;    COLLINS, 

KEESE  &  Co,  New-York.— PERKINS  &  MARVIN,  Boston.— 

THOMAS,  COWPERTHWAIT  &  Co,  Philadelphia.— 

GUSHING  &  SONS,  Baltimore. 

1839. 


DAVIES'  COURSE  OF  MATHEMATICS. 


DAVIES'  MENTAL  and  PRACTICAL  ARITHMETIC— Designed 
for  the  use  of  Academies  and  Schools.  It  is  the  purpose  of  this 
work  to  explain,  in  a  brief  and  clear  manner,  the  properties  of  numbers, 
and  the  best  rules  in  their  various  applications. 

DAVIES'  KEY— To  Mental  and  Practical  Arithmetic. 

DAVIES'  FIRST  LESSONS  in  ALGEBRA— Being  an  introduction 
to  the  Science. 

DAVIES'  BOURDON'S  ALGEBRA— Being  an  abridgment  of  the 
work  of  M.  Bourdon,  witli  the  addition  of  practical  examples. 

DAVIES'  LEGENDRE'S  GEOMETRY  and  TRIGONOMETRY 
— Being  an  abridgment  of  the  work  of  M.  Legendre  with  the  addition 
of  a  treatise  on  Mensuration  of  Planes  and  Solids,  and  a  table  of 
Logarithms  and  Logarithmic  sines. 

DAVIES'  SURVEYING— With  a  description  and  plates  of,  the  Theo- 
dolite.  Compass,  Plane-Table  and  Level, — also.  Maps  of  the  Topo- 
graphical Signs  adopted  by  the  Engineer  Department,  and  an  explana- 
tion of  the  method  of  surveying  the  Public  Lands. 

DAVIES'  ANALYTICAL  GEOMETRY— Embracing  the  Equations 
of  the  Point  and  Straight  Line — of  the  Conic  Sections — of  the  Line 
and  Plane  in  Space — also,  the  discussion  of  the  General  Equation 
of  the  second  degree,  and  of  surfaces  of  the  second  order. 

DAVIES'  DESCRIPTIVE  GEOMETRY— V^^ith  its  applications  to 
Spherical  Projections. 

DAVIES'  SHAD0V7S  AND  LINEAR  PERSPECTIVE. 

DAVIES'  DIFFERENTIAL  AND  INTEGRAL  CALCULUS— 
With  numerous  applications. 


Entered  according  to  the  Act  of  Congress,  in  the  year  one  thousand 
eight  hundred  and  thirty-eight,  by  Charles  Davies,  in  the  Clerk's 
Office  of  the  District  Court  of  the  United  States,  for  the  Southern 
District  of  New  York. 


Stereotyped  by  Henry  W.  Rees, 
32  Ann  Street,  New  York. 


PREFACE. 


Although  Algebra  naturally  follows  Arithmetic  in  a 
course  of  scientific  studies,  yet  the  change  from  num- 
bers to  a  system  of  reasoning  entirely  conducted  by 
letters  and  signs  is  rather  abrupt  and  not  unfrequently 
discourages  and  disgusts  the  pupil. 

In  the  First  Lessons  it  has  been  the  intention  to 
form  a  connecting  link  between  Arithmetic  and  Algebra, 
to  unite  and  blend,  as  far  as  possible,  the  reasoning  on 
numbers  with  the  more  abstruse  method  of  analysis. 

The  Algebra  of  M.  Bourdon  has  been  closely  fol- 
lowed. Indeed,  it  has  been  a  part  of  the  plan,  to  furnish 
an  introduction  to  that  admirable  treatise,  which  is  justly 
considered,  both  in  Europe  and  this  country,  as  the  best 
work  on  the  subject  of  which  it  treats,  that  has  yet 
appeared. 

111869 


4  PREFACE. 

This  work,  however,  even  in  its  abridged  form,  is  too 
voluminous  for  schools,  and  the  reasoning  is  too  elaborate 
and  metaphysical  for  beginners. 

It  has  been  thought  that  a  work  which  should  so  far 
modify  the  system  of  Bourdon  as  to  bring  it  within  the 
scope  of  our  common  schools,  by  giving  to  it  a  more 
practical  and  tangible  form,  could  not  fail  to  be  useful. 
Such  is  the  object  of  the  First  Lessons.  It  is  hoped 
they  may  advance  the  cause  of  education,  and  prove 
a  useful  introduction  to  a  full  course  of  mathematical 
studies. 

HiRTFORD,  September^  1838. 


CONTENTS. 


CHAPTER  I. 

PRELIMINARY   DEFINITIONS    AND    REMARKS. 

ARTICLESf! 

Algebra — Definitions — Explanation  of  the  Algebraic  Signs,      -       1 — 23 

Similar  Terms — Reduction  of  Similar  Terms,         -        -        •  23 — 26 

Addition— Rule,  -----.-.  26 — 28 

Subtraction — Rule — Remark,        -         -        -        -        -        -  28 — 33 

Multiplication — Rule  for  Monomials,      -----  33 — 36 

Rule  for  Polynomials  and  Signs,   ------  36 — 38 

Remarks — Properties  Proved,        ------  38 — 43 

Division  of  Monomials — Rule,       ------  42 — 45 

Signification  of  the  Symbol  ao,       -        -        ^        -        -        -  45 — 46 

Of  the  Signs  in  Divison,        -------  46 — 47 

Division  of  Polynomials,       ------  47_49 

CHAPTER  H. 


ALGEBRAIC   FRACTIONS. 

l)efinitions — Entire  Quantity — Mixed  Quantity,       •  -  49—53 

To  Reduce  a  Fraction  to  its  Simplest  Terms          -  -  .  62 

To  Reduce  a  Mixed  Quantity  to  a  Fraction,            -  -  -  63 

To  Reduce  a  Fraction  to  an  Entire  or  Mixed  Quantity,  -  -  64 

To  Reduce  Fractions  to  a  Common  Denominator,  -  -  -  65 

To  Add  Fractions,       •        - 56 


6 


CONtENTS. 


To  Subtract  Fractions, 
To  Multiply  Fractions, 
To  Divide  Fractions,    - 


ARTICLES. 
57 


59 


CHAPTER  III 


EQUATIONS    OP    THE   FIRST    DEGREE. 


Definition  of  an  Equation — Properties  of  Equations, 
Transformation  of  Equations — First  and  Second,    -         -         - 
Resolution  of  Equations  of  the  First  Degree — Rule, 
Questions  involving  Equations  of  the  First  Degree, 
Equations   of    the    First    Degree   involving   Two   Unknown 

Quantities,  -         -         -        -         -         -         -- 

Elimination — By  Addition — By  Subtraction — By  Comparison,  - 
Resolution  of  Questions   involving  Two  or   more  Unknown 

Quantities,  --------- 


60- 

-66 

66- 

-70 

70 

71- 

-73 

72 

73- 

-76 

76—79 


CHAPTER  IV. 

OF    POWERS. 

Definition  of  Powers,   --------  79 

To  raise  Monomials  to  any  Power,        -----  80 

To  raise  Polynomials  to  any  Power,      -----  81 

To  raise  a  Fraction  to  any  Power,         -----  82 — 83 

Binomial  Theorem,      --------  84 — 90 


CHAPTER  V. 

Definition    of   Squares — Of    Square    Roots — And    Perfect 

Squares, 90—96 

Rule  for  Extracting  the  Square  Root  of  Numbers,       -        -  96 — 100 

Square  Roots  of  Fractions,        ------  lOO — 103 

Square  Roots  of  Monomials, 103 — 107 

Calculus  of  Radicals  of  the  Second  Degree,        -        -        •  107—109 


CONTENTS.  7 

ARTICLES. 

Addition  of  Radicals,         .---.•-  109 

Subtraction  of  Radicals,    -         -         -         -         -         -         -  110 

Multiplication  of  Radicals,          -         -         -         -         -         -  111 

Division  of  Radicals,         -         -         -         -         -         -         -  112 

Extraction  of  the  Square  Root  of  Polynomials,    -         .         -  113 — 116 

CHAPTER  VI. 

Equations  of  the  Second  Degree,        -         -         -         y         -  116 

Definition  and  Form  of  Equations,      -         -         -         -  116 — 118 

Incomplete  Equations,       -------  us — 122 

Complete  Equations,         -------  122 

Four  Forms, 123—127 

Resolution  of  Equations  of  the  Second  Degree,  -        -         -  127 — 128 

Properties  of  the  Roots,    -         -        -                  -         .        -  128—134 

CHAPTER  VIL 

Of  Progressions,       ------.-•  135 

Progressions  by  Differences,      ------  136 — 138 

Last  Term, 138—140 

Sum  of  the  Extremes — Sum  of  the  Series,           -       '  -         -  140 — 141 

The  Five  Numbers — To  Find  any  Number  of  Means,          -  141 — 144 

Geometrical  Proportion  and  Progression,    -         -         -         -  144 

Various  Kinds  of  Proportion,     ------  144 — 166 

Geometrical  Progression, --  166 

Last  Term— Sum  of  the  Series, 167 — 171 

Progressions  having  an  Infinite  Number  of  Terms,    '  -        -  171 — 172 

The  Five  Numbers— To  Find  One  Mean  -         -        -        -  172—173 


FIRST  LESSONS 

IN 

ALGEBRA 


CHAPTER  I. 

Preliminary  Definitions  and  Remarks. 

1.  Quantity  is  a  general  term  embracing  every  thing 
which  can  be  increased  or  diminished.  , 

2.  Mathematics  is  the  science  of  quantity. 

3.  Algebra  is  that  branch  of  mathematics  in  which  the 
quantities  considered  are  represented  by  letters,  and  the  ope- 
rations to  be  performed  upon  them  are  indicated  by  signs. 
These  letters  and  signs  are  called  symbols. 

4.  The  sign  +,  is  called  plus  ;  and  indicates  the  addition 
of  two  or  more  quantities.  Thus,  9  +  5,  is  read,  9  plus  5, 
or  9  augmented  by  5. 

If  we  represent  the  number  nine,  by  the  letter  «,  and 
the  number  5  by  the  letter  h,  we  shall  have  a+ 5,  which  is 
read,  a  plus  h  ;  and  denotes  that  the  number  represented  by 
a  is  to  be  added  to  the  number  represented  by  b. 

5.  The  sign—,  is  called  minus ;  and  indicates  that  one 


Quest. — 1.  What  is  quantity  1  2.  What  is  Mathematics  1  3.  What 
is  Algebra  1  "WHiat  are  these  letters  and  signs  called  1  4.  What  does  the 
sign  plus  indicate  1     6.  What  does  the  sign  minus  indicate  ? 


10  FIRST    LESSONS    IN    ALGEBRA. 

quantity  is  to  be  subtracted  from  another.     Thus,  9— 5  is 
read,  9  minus  5,  or  9  diminished  by  5. 

In  like  manner,  a—b,  is  read,  a  minus  h,  or  a  diminished 
by  h. 

6.  The  sign  x  ,  is  called  the  sign  of  multiplication ;  and 
when  placed  between  two  quantities,  it  denotes  that  they 
are  to  be  multiplied  together.  The  multiplication  of  two 
quantities  is  also  frequently  indicated  by  simply  placing  a 
point  between  them.  Thus,  36  x  25,  or  36.25,  is  read,  36 
multiplied  by  25,  or  the  product  36  by  25. 

7.  The  multiplication  of  quantities,  which  are  represented 
by  letters,  is  indicated  by  simply  writing  them  one  after  the 
other,  without  interposing  any  sign. 

Thus,  ah  signifies  the  same  thing  as  axh,  or  as  a,h ; 
and  ahc  the  same  as  axhxc,  or  as  a.b.c.  Thus,  if  we 
suppose  a=:36,  and  Z>r=25,  we  have 

a5  =  36x25=:  900. 

Again,  if  we  suppose  ^=2,  b=3  and  c=4,  we  have 

ahc=z2x3x4=z24. 

It  is  most  convenient  to  arrange  the  letters  of  a  product 
in  alphabetical  order. 

8.  In  the  product  of  several  letters,  as  abc,  the  single  let- 
ters, a,  bj  and  c,  are  called  factors  of  the  product.  Thus, 
in  the  product  ah,  there  are  two  factors,  a  and  b ;  in  the 
product  abc,  there  are  three,  a,  5,  and  c. 


Quest. — 6.  What  is  the  sign  of  multiplication  1  What  does  the  sign 
of  multiphcation  indicate  1  In  how  many  ways  may  multiplication  be 
expressed  1  7.  If  letters  only  are  used,  how  may  their  multiplication  be 
expressed!  8.  In  the  product  of  several  letters,  what  is  each  letter 
called  1     How  many  factors  in  db  1 — In  abc  ? — In  abed  1 — In  ahcdfl 


DEFINITION    OF    TERMS.  11 

9.  There  are  three  signs  used  to  denote  division.     Thus, 

a-^b  denotes  that  a  is  to  be  divided  by  h. 

-        denotes  that  a  is  to  be  divided  by  h. 
o 

a\b      denotes  that  a  is  to  be  divided  by  b. 

10.  The  sign  =,  is  called  the  sign  of  equality^  and  is 
read,  is  equal  to.  When  placed  between  two  quantities,  it 
denotes  that  they  are  equal  to  each  other.  Thus,  9  — 5=4  : 
that  is,  9  minus  5  is  equal  to  4  :  Also,  a-\-bz:z.c,  denotes  that 
the  sum  of  the  quantities  a  and  b  is  equal  to  c. 

If  we  suppose  a  =  10,  and  b  =  5,  we  have 

a-{-b=zCj     and     10  +  5=c  =  15. 

1 1.  The  sign  >,  is  called  the  sign  of  inequality,  and  is 
used  to  express  that  one  quantity  is  greater  or  less  than 
another. 

Thus,  « >  Z>  is  read,  a  greater  than  b ;  and  a<^b  is 
read,  a  less  than  b  ;  that  is,  the  opening  of  the  sign  is  turned 
towards  the  greater  quantity.  Thus,  if  a =9,  and  5=4,  we 
write,  9>4. 

12.  If  a  quantity  is  added  to  itself  several  times  as 
a-\-a+a-\-a+a-\-a,  we  generally  write  it  but  once,  and 
then  place  a  number  before  it  to  show  how  many  times  it 
is  taken.     Thus, 

a-\-a-\-a-\-a-{-a=i^a. 


Quest. — 9.  How  many  signs  are  used  in  division  1  What  are  they  1 
10.  What  is  the  sign  equality  \  When  placed  between  two  quantities, 
what  does  it  indicate!  11.  For  what  is  the  sign  of  inequahty  usedl 
Which  quantity  is  placed  on  the  side  of  the  opening "?  12.  What  is  a  co- 
efficient 1  How  many  times  is  ah  taken  in  the  expression  ah  ]  In  3ai ! 
In  ^ah  1  In  bah  1  In  6a5  \  If  no  co-efficient  is  written,  what  co-efficient 
is  understood'^ 


12  FIRST    LESSONS    IN    ALGEBRA. 

The  number  5  is  called  the  co-efficient  of  a,  and  denotes 
that  a  is  taken  5  times. 

When  the  co-efficient  is  1  it  is  generally  omitted.  Thus, 
a  and  \a  are  the  same,  each  being  equal  to  «,  or  to  one  a. 

13.  If  a  quantity  be  multiplied  continually  by  itself,  as 
axaxaxaxa,  we  generally  express  the  product  by  writing 
the  letter  once,  and  placing  a  number  to  the  right  of,  and  a 
little  above  it :  thus, 

aXaXaxa  Xa=:a^. 

The  number  5  is  called  the  exponent  of  a,  and  denotes 
the  number  of  times  which  a  enters  into  the  product  as  a 
factor.  For  example,  if  we  have  a^,  and  suppose  a  =  3, 
we  write, 

a^  =  axaXa=3^=:3xSx3=z27. 
If  a=i4,  a3  =  43=:4x4x4  =  64, 

and  for  ariz5,        a^=5^=:5x5x 5  =  125. 

If  the  exponent  is  1  it  is  generally  omitted.  Thus,  a^  is 
the  same  as  a,  each  expressing  a  to  the  first  power. 

1 4.  The  power  of  a  quantity  is  the  product  which  results 
from  multiplying  the  quantity  by  itself.    Thus,  in  the  example 

a3— 43  =  4x4x4  =  64, 

64  is  the  third  power  of  4,  and  the  exponent  3  shows  the 
degree  of  the  power. 

15.  The  sign  V     ,  is  called  the  radical  sign,  and  when 


Quest. — 13.  What  does  the  exponent  of  a  letter  showl  How  many 
times  is  a  a  factor  in  a2 1  \n  a^\  In  a^  1  In  as '?  If  no  exponent  is 
written,  what  exponent  is  understood  1  14.  What  is  the  power  of  a 
quantity  '\  What  is  the  third  power  of  2  ]  Express  the  4th  power  of  a, 
15.  Express  the  square  root  of  a  quantity.  Also  the  cube  root.  Also 
the  4th  root. 


DEFINITION    OP   TERMS.  13 

prefixed  to  a  quantity,  indicates  that  its  root  is  to  be  ex- 
tracted.    Thus, 

'^oTov  simply  -/tTdenotes  the  square  root  of  a. 

-y/o'denotes  the  cube  root  of  a, 

-t/o"  denotes  the  fourth  root  of  a. 


The  number  placed  over  the  radical  sign  is  called  the  in- 
dex of  the  root.  Thus,  2  is  the  index  of  the  square  root,  3 
of  the  cube  root,  4  of  the  fourth  root,  &;c. 

If  we  suppose    a  =  64,    we  have 

16.  Every  quantity  written  in  algebraic  language,  that 
is,  with  the  aid  of  letters  and  signs,  is  called  an  algebraic 
quantity,  or  the  algebraic  expression  of  a  quantity.     Thus, 

is  the  algebraic  expression  of  three  times 
the  number  a ; 
-25^^  ^^®  algebraic  expression  of  five  times 
(      the  square  of  a  ; 


3     i 
ia  < 


r  is  the  algebraic  expression  of  seven  times 

7a^'^  ^      the  product  of  the  cube  of  a  by  the  square 

(      of  6; 

q  _ci  (  is  the  algebraic  expression  of  the  difference 

I      between  three  times  a  and  five  times  b  ; 

2a2  — 3a5+452^ 


is  the  algebraic  expression  of  twice  the 
square  of  «,  diminished  by  three  times 
the  product  of  a  by  5,  augmented  by  four 
times  the  square  of  b. 

1.  Write  three  times  the  square  of  a  multiplied  by  the 
cube  of  b.  Ans.  3a^\ 


Quest. — 16.  What  is  an  algebraic  quantity  1     Is  5ab  an  algebraic 
quantity  1     Is  9a  1     Is^l     IsZb  —  x] 

2 


14  FIRST    LESSONS    IN    ALGEBRA. 

2.  Write  nine  times  the  cube  of  a  multiplied  by  h,  dimin- 
ished by  the  square  of  c  multiplied  by  d.      Ans,  ^a%—c^d. 

3.  If  «=i2,  Z>iiz3,  and  cz=zb,  what  will  be  the  value  of 
3^2  multiplied  by  V^  diminished  by  a  multiplied  by  h  multi- 
plied by  c.     We  have 

^a%'^—ahc  =  ^  X  22  X  32-2  X  3  X  5=78. 

4.  If  cz=4,  5  =  6,  c=7,  cZ=8,  what  is  the  value  of 
9a?+hc—ad1  Ans.   154. 

5.  If  fl=7,  Z>=3,  c=7,  (?=1,  what  is  the  value  of 
Gad-^W^c  —  Ad'^'^.  Ans,  227. 

6.  If  a  =  5,  5  =  6,  c=6,  t?=5,  what  is  the  value  of 
9ahc  —  Sad-{-Ahc'i  Ans.   1564. 

7.  Write  ten  times  the  square  of  a  into  the  cube  of  h  into 
c  square  into  d^, 

17.  When  an  algebraic  quantity  is  not  connected  with 
any  other,  by  the  sign  of  addition  or  subtraction,  it  is  called 
a  monomial^  or  a  quantity  composed  of  a  single  term,  or  sim- 
ply, a  term.     Thus, 

3«,     5a2,     7a352^ 

are  monomials,  or  single  terms. 

18.  An  algebraic  expression  composed  of  two  or  more 
parts,  separated  by  the  sign  +  or  — ,  is  called  d,  polynomial^ 
or  quantity  involving  two  or  more  terms.     For  example, 

3a-55     and     2a^—'ich+W^ 

are  polynomials. 

19.  A  polynomial  composed  of  two  terms,  is  called  a  hi" 
nomial ;  and  a  polynomial  of  three  terms  is  called  a  trinomial. 

Quest. — 17.  What  is  a  monomial  1  Is  ^ab  a  monomial  1  18.  What 
is  a  polynomial  1  Is  3a  — &  a  polynomial  1  19.  What  is  a  binomial] 
What  is  a  trinomiall 


DEFINITION    OF   TERMS.  15 

20.  Each  of  the  literal  factors  which  compose  a  term  is 
called  a  dimension  of  this  term :  and  the  degree  of  a  term  is 
the  number  of  these  factors  or  dimensions.     Thus, 

1  is  a  term  of  one  dimension,  or  of  the  first 


IS 

second  degree. 


degree. 

{ 


-  ,  ^  is  a  term  of  two  dimensions,  or  of  the 


7  3^  2  —  7      h     (  is   of  six   dimensions,   or   of  the   sixth 
"~  I      degree. 

2 1 .  A  polynomial  is  said  to  be  homogeneous,  when  all 
its  terms  are  of  the  same  degree.     The  polynomial 

^a—2h+c  is  of  the  first  degree  and  homogeneous. 

—Aah-\-h'^  is  of  the  second  degree  and  homogeneous. 

^a^c—4:c'^+2cH  is  of  the  third  degree  and  homogeneous. 

8a3+4«6+c  is  not  homogeneous. 

22.  A  vinculum  or  bar   — ,  or  a  parenthesis  ( ), 

is  used  to  express  that  all  the  terms  of  a  polynomial  are  to 
be  considered  together.     Thus, 

a+h-fcxh,    or    {a+h+c)xh, 

denotes  that  the  trinomial  a+h+c  is  to  be  multiplied  by  b  ; 

also,       a-\-b+cXc-\-d-\-f,   or    {a+h+c)x{c-\-d+f), 

denotes  that  the  trinomial   a+h+c   is  to  be  multiplied  by 
the  trinomial    c+d+f. 

When  the  parenthesis  is  used,  the  sign  of  multiplication 
is  usually  omitted.     Thus, 

(a+h+c)y.h   is  the  same  as    [a+h  +  c)h. 

Quest. — 20.  What  is  the  dimension  of  a  termi  What  is  the  degree 
of  a  term  1  How  many  factors  in  ^ahc  1  Which  are  they  1  What 
is  its  degree "!  21.  When  is  a  polynomial  homogeneous  1  Is  the  polyno- 
mial 2a36+3a262  homogeneous'?  Is  ^a^h  —  ^3]  22.  For  what  is  the 
vinculum  or  bar  used  1     Can  you  express  the  same  with  the  parenthesis  1 


16  FIRST  LESSONS  IN  ALGEBRA. 

!23.  The  terms  of  a  polynomial  which  are  composed  of 
the  same  letters,  the  same  letters  in  each  being  affected 
with  like  exponents,  are  called  similar  terms. 

Thus,  in  the  polynomial 

lah + ^ah —AaW + 5a^% 

the  terms  7ab,  and  Sab,  are  similar  :  and  so  also  are  the 
terms  —4a^^  and  da^b"^,  the  letters  and  exponents  in  both 
being  the  same.  But  in  the  binomial  8a^-\-7ab^,  the 
terms  are  not  similar ;  for,  although  they  are  composed  of 
the  same  letters,  yet  the  same  letters  are  not  affected  with 
like  exponents, 

24.  When   an   algebraic    expression   contains   similar* 
terms,  it  may  be  reduced  to  a  simpler  form. 

1.  Take  the  expression  3ab  +  2aby  which  is  evidently 
equal  to    5ab. 

2.  Reduce  the  expression  3ac-\-9ac-\-2ac  to  its  simplest 
form.  Ans.  14ac. 

3.  Reduce  the  expression  abc-\-4abc-\-5abc  to  its  sim- 
plest form. 

In  adding  similar  terms  together  we  abc 

take  the  sum  of  the   coefficients  and  4abc 

annex  the  literal  part.     The  first  term,  5abc 

abc,    has    a   coefficient    1    understood,  lOabc 
(Art.  12). 

25.  Of  the  different  terms  which  compose  a  polynomial, 
some  are  preceded  by  the  sign  +,  and  the  others  by  the 
sign  — ,  The  first  are  called  additive  terms,  the  others, 
subtractive  terms. 


Quest. — 23.  What  are  similar  terms  of  a  polynomiall  Are  Sasja 
and  6a2&2  similar  1  Are  2a2i2  and  2a3j2  ]  24.  If  the  terms  are  positive 
and  similar,  may  they  be  reduced  to  a  shnplerformt    In  what  way' 


DEFINITION    OF    TERMS.  17 

The  first  term  of  a  polynomial  is  commonly  not  preceded 
by  any  sign,  but  then  it  is  understood  to  be  affected  with  the 
sign  +. 

1 .  John  has  20  apples  and  gives  5  to  William :  how 
many  has  he  left  ? 

Now,  let  us  represent  the  number  of  apples  which  John 
has  by  a,  and  the  number  given  away  by  h  :  the  number  he 
would  have  left  would  then  be  represented  by    a—b. 

2.  A  merchant  goes  into  trade  with  a  certain  sum  of 
money,  say   a   dollars  ;  at  the  end  of  a  certain  time  he  has    / 
gained   b   dollars  :  how  much  will  he  then  have  ? 

Ans.  a-{-b  dollars. 

If  instead  of  gaining  he  had  lost  b  dollars,  how  much 
would  he  have  had  1  Ans.  a  —  b  dollars. 

Now,  if  the  losses  exceed  the  amount  with  which  he 
began  business,  that  is,  if  b  were  greater  than  a,  we  must 
prefix  the  minus  sign  to  the  remainder  to  show  that  the 
quantity  to  be  subtracted  was  the  greatest. 

Thus,  if  he  commenced  business  with  $2000,  and  lost 
$3000,  the  true  difference  would  be  —1000  :  that  is,  the 
subtractive  quantity  exceeds  the  additive  by  $fOOO. 

3.  Let  a  merchant  call  the  debts  due  him  additive,  and 
the  debts  he  owes  subtractive.  Now,  if  he  has  due  him 
$600  from  one  man,  $800  from  another,  $300  from  another, 
and  owes  $500  to  one,  $200  to  a  second,  and  $50  to  a 
third,  how  will  the  account  stand?        Ans.  $950  due  him. 

4-.  Reduce  to  its  simplest  form  the  expression 
3a^  -{-5a'^b  —  3a^  +  4a^ — Ga^J  _  a^, 

Quest. — 25.  What  are  the  terms  called  which  are  preceded  by  the 
sign  +  ?  What  are  the  terms  called  which  are  preceded  by  the  sign  — . 
If  no  sign  is  prefixed  to  a  term,  what  sign  is  understood  ?  If  some  of  the 
terms  are  additive  and  some  subtractive,  may  they  be  reduced  if  similar  ? 
Give  the  rule  for  reducing,  them, .  Does  the  reduction  affect  the  expo- 
nents, or  only  the  coefficients  ? 

2* 


18  FIRST  LESSONS  IN  ALGEBRA. 

Additive  terms.  Suhtractive  terms. 

+   ^a?h  —   Za^h 

+   ba%  —   Qa?h 

4-   ^a^h  —     a^h 

Sum  +\2a%  Sum  — lOo^^'. 

But,  \2a%-\0a?h=z2a?b. 

Hence,  for  the  reduction  of  the  similar  terms  of  a  polyno- 
mial we  have  the  following 

RULE. 

I.  Form  a  single  additive  term  of  all  the  terms  preceded  hy 
the  sign  plus  ;  this  is  done  by  adding  together  the  coefficients 
of  those  terms ^  and  annexing  to  their  sum  the  literal  part. 

II.  Form,  in  the  same  manner,  a  single  suhtr active  term. 

III.  Subtract  the  less  sum  from  the  greater,  and  prefix  to 
the  result  the  sign  of  the  greater. 

Remark. — It  should  be  observed  that  the  reduction  affects 
only  coefficients,  and  not  the  exponents. 

•  EXAMPLES. 

1.  Reduce  to  its  simplest  form  the  polynomial 

-\-2a%c'^  —  A  a^c^ + Sa^c^- — 8a^c^  + 1 1  a^c'^. 

Find  the  sum  of  the  additive  and  subtractive  terms  sepa- 
rately, and  take  their  difference  :  thus. 

Additive  terTns.  Subtractive  terms. 

+  2a^c'^  —  4a^c^ 

+   6a^c^  '  —   Sa^c'^ 

+  lla^  Sum  —I2a^c^ 
Sum  -\-l9o^ 

Hence,  the  given  polynomial  reduces  to 
1 9a^bc^ — 1 2u^c^ = 7a^bc^. 


ADDITION.  19 

2.  Reduce  the  polynomial  4a26— Sa^i— Oa^^+lla^j  to 
its  simplest  form.  Ans.   —2a^b, 

3.  Reduce  the  polynomial  7abc'^—abc'^—7abc^-\-8abc^ 
+  6abc'^  to  its  simplest  form.  Ans.       ISabc^. 

4.  Reduce  the  polynomial  9cb^—Sac^-\-l5cb^-\-8ca-{-9ac'^ 
— 24cb^  to  its  simplest  form.  Ans.  ac^-^-Sca. 

The  reduction  of  similar  terms  is  an  operation  peculiar  to 
algebra.  Such  reductions  are  constantly  made  in  Algebraic 
Addition,  Subtraction,  Multiplication,  and  Division. 


ADDITION. 

26.  Addition  in  Algebra,  consists  in  finding  the  simplest 
equivalent  expression  for  several  algebraic  quantities,  con- 
nected together  by  the  sign  plus  or  minus.  Such  equivalent 
expression  is  called  their  sum. 

1.  What  is  the  sum  of 

3ax-\-2ab    and    —2ax  +  ab.    — ^- 

3ax~{-2ab 

We  reduce  the  terms  as  in  Art.  25,  —2ax-\-  ab 

and  find  for  the  sum ax-{-3ab 

2.  Let  it  be  required  to  add  together  \  -, 
the  expressions  :                                          i 


The  result   is 3a-\-bb-\-2c 

an  expression  which  cannot  be  reduced  to  a  more  simple 
form. 


Quest. — 26.  What  is  addition  in  Algebra  ?     What  is  such  simplest 
and  oquivalent  expression  called  ? 


^0  FIRST  LESSONS  IN  ALGEBRA. 

Again,  add  together  the   monomials  ?  2a^P 


The  result  after  reducing  (Art.  25),  is  .  .  ISa^b^ 

r    2a^-'Aah 
3.  Let  it  be  required  to  find  the  sum  \    q2_QA4_7,a 


of  the  expressions 


2ah~b}p' 


Their  sum,  after  reducing  (Art.  25)  is  .  ba^—bah—Al)^ 

27.  As  a  course  of  reasoning  similar  to  the  above 
would  apply  to  all  polynomials,  we  deduce  for  the  addition 
of  algebraic  quantities  the  following  general 

RULE. 

I.  Write  down  the  quantities  to  he  added  so  that  the  similar 
terms  shall  fall  under  each  other,  and  give  to  each  term  its 
proper  sign. 

II.  Reduce  the  similar  terms,  and  annex  to  the  results  the 
terms  which  cannot  he  reduced,  giving  to  each  term  its  respec* 
tive  sign. 

EXAMPLES. 

1.  What  is  the  sum  of  Sax,  5ax,   ■—2ax  and  ISax.  ? 

A?is.  I9ax. 

2.  What  is  the  sum  of  4ah-\-8ac  and  2a5— 7ac+<^.  ? 

Ans.  6ah-}-ac-{-d. 

3.  Add  together  the  polynomials, 

3a'^  —  2h^—iah,  5a'^  —  h^-\-2ah,  and  3ah-^3c^^2h^. 
The  term  3a^  being  similar  to 
5a2,    we  write    Sa^   for  the  result 
of  the  reduction  of  these  two  terms,  ^ 
at  the  same  time  slightly  crossing 
them,  as  in  the  first  term. 

QuBST. — 37.  Give  the  rule  for  the  addition  of  Algebraic  quantities. 


-\-3(f^h—2b^—3c^ 
8a24.   ab  —  5b^—3c^ 


ADDITION.  21 

Passing  then  to  the  term  —4ah,  which  is  similar  to-\-2ab 
and  -{-Sab,  the  three  reduce  to  -\-ab,  which  is  placed  after 
8a^,  and  the  terms  crossed  like  the  first  term.  Passing 
then  to  the  terms  involving  b^,  we  find  their  smn  to  be 
— 5b'^,    after  which  we  write    — Sc^. 

The  marks  are  drawn  across  the  terms,  that  none  of  them 
may  be  overlooked  and  omitted. 

(4)  (5)  (6) 

a  6a  5a 

a  ba  5b 

2a  Tla  5a-\-5b 


(7) 

Sab 

(8) 

Sac 

5ab 
Sab 

Sac 
llac 

(11) 
12a—  6c 

—  3a—   9c 

(9)  (10) 

7abc-{-9ax  Sax-{-Sb 

—  Sabc—Sax  5ax — 95 

Aabc-\-6ax  \Sax — 6b                     9a — 15c 

Note. — If  az=:5,  5=4,  c=:2,  a?=l,  what  are  the  values 
of  the  several  sums  above  found. 


(12) 

(13 

) 

(14) 

9a+/ 

6ax — 

8ac 

3«/+  g  -\-m 

6a-\-g 

-7ax— 

9ac 

ag—Saf—m 

2a-/ 

ax -\- 17  ac 

ab—  ag-\-Sg 

^-\-g 

0 

0 

ab-{-4g 

(15) 

(16) 

7x-{-Sab+ 

3c 

8a;2+ 

9acx-\-lSa%'^c^ 

Sx  —  Sab  — 

5c 

—7x^— 

ISacx+lAa^b'^c'^ 

5x~9ab  — 

9c 

-4a;2+ 

4acx—20aH^c^ 

9x—9ab  —  llc  —  3a;^+   0       +    7a^'^c^ 

(17)  (18) 

22A-3c-7/+3^  l9ah^-\-Sa^*-Sax^ 

—   3A+8c— 2/— 9^4-  5^  —  17aA2— 9a35^  +  9ax^ 

19A4  5c-9/'-6^+5^  2ah^—6a^^-\-  ax^ 


29  FIRST  LESSONS  IN  ALGEBRA. 

(19)  (20) 

7a?— 9y+5;^+3—  g  Sa+  h 

—  0?  — 3y          —8—-  ^  2a—   5+   c 

—  a;+  y—^z+\  +  lg  — 3«+   J          +2df 
—2a?+6y+ 3-^—1—  g  —6J— 3c+-3<? 

y+8y— 5^+9+  ^  —5a         +7c— 8(Z 

4a:+3y+0  +4+5^  2a— 554-5c— 3J 

21.  Add  together  — Z»+3c— <Z— 115e+6/— 5^,  35— 2(r 
— Sf^— g_|-27/,  5c— 8J+3/— 7^,  ~73»— 6c+17(Z+9e-5/ 
+  11^,    — 35-5J— 2e+6/— 9^+A. 

Ati^.  — 85— 109e+37/— 10^+^. 

22.  Add  together  the  polynomials  7a^5— 3a5c— 852c— 9c3 
+  cJ2,  8a5c— 5a25-|-3c3— 452c  +  c(i2  and  4a25_8c3+9^^c 
— 3£?3.  ^n^.  6a25  +  5a5c— 352c— 14c3+2ccZ2— 3cZ3. 

23.  What  is  the  sum  of  5a25c+65a;'-4a/,  — 3a25c— 65ar 
+  14ft/,    — a/+95a;+2a25c,    +6a/— 85ir+6a25c. 

Ans,   \Oa?'hc-\-hx-{-l^af, 

24.  What  is  the  sum  of  a^n^+Za^m  +  h,  —6a'^n'^—6a^m—h, 
+  95— 9a3m— 5a2n2.  Ans,  -^lOaV—Ua^m+Ob 

25.  What  is  the  sum  of  4a352c — 16a^a: — 9ax^d,  +6a352c 

Ans.  a^b^c-{-ax^d, 

26.  What  is  the  sum  of  —7^+35+4^—25,  +3^ 
—35+25.  Ans,  0. 

27.  What  is  the  sum  of  a5  +  3a:y— m— n,  —  6a:y— 3m 
+  lln+ctZ,    +3a?y  +  4m — lOn+fg.  Ans.  ah-\-cd-{-fg. 

28.  What  is  the  sum  of  4iry+n  +  6aa;+9am,  — 6a:y+6;i 
^Qax — 8am,   2xy — In-^-ax — am,  Ans.  -\-ax^ 


SUBTRACTION.  23 

29.  Add  the  polynomials  I9a^x^--I2a^cb,  5a^x^  +  l4a^cb 
--lOax,    ^2a^x^'-l2a^cb,  and  —ISa^x^  —  Ua^cb-^-dax. 

Ans,  Aa^x^b—^^a^cb—ax. 

30.  Add   together    2a+b+c,    5a+25+3ac,    a+c+ac, 
and — 3a — 9ac — 8^.  Ans.  6a—5b-{-2c—5ac. 

31.  Add  together     .     ba^b+dcx+dbc^,    7cx—8a'^b,    and 
—  I5cx—9bc^-^2a'^b.  Ans.   —aF'b  —  2cx. 

32.  Add  together  Qax+^ab+3a^b'^c",  —ISax+Qa^+lOab 
and  lOax—l^ab—Qa^bH'^.  Ans.   —3a'^b'^c'^+6a^. 

33.  Add  together  3a^-{-5a^^c'^'-9a^x,  la^—aa^^c^—lOa^x 
Bind  I0ab+iea^b^c^+l9a^x.       Ans.  I0a^  +  l3a^^c^+10ab. 


L 


SUBTRACTION. 


28.  Subtraction,  in  Algebra,  consists  in  finding  the  sim- 
plest expression  for  the  difference  between  two  algebraic 
quantities. 

Thus,  the  difference  between  6a  and  3a  is  expressed  bv 

6a — 3a=3a ; 
and  the  difference  between  7a^  and  3a^  by 
7a^-3a^  =  4a^. 

In  like  manner  the  difference  between  4a  and  35  is 
expressed  by  4a— 3b. 

Hence,  If  the  quantities  are  similar,  subtract  the  coefficients ; 
and  if  they  are  not  similar,  place  the  minus  sign  before  the 
quantity  to  be  subtracted. 

Quest. — 28.  In  what  does  subtraction  in  Algebra  consist?  How  do 
you  find  this  difference  when  the  quantities  are  positive  and  similar  ? 
When  they  are  not  similar,  how  do  you  express  the  difference  ? 


24 


I 

FIRST    LESSONS    IN    ALGEBRA. 

(1) 

(2) 

(3) 

From 

Sab 

6ax 

9abc 

take 

2ab 

Sax 

7abc 

Rem. 

ah 

Sax 

2ahc. 

(4) 

(5) 

(6) 

From 

16«252c 

\7a%^c       ' 

24^2^,2^ 

take 

9a%'^c 

Sa-'h^c 

7a^^x 

Rem. 

laWc 

\Aa%'^c 

\la%H 

(7) 

(8) 

(9) 

From 

Sax 

Aahx 

2  am 

take 

8c 
3«a;— 8c 

9ac 
4abx^9ac 

ax 

Rem. 

2am— ax. 

29.  Let  it  be  required  to  subtract  from     Aa 

the  binomial 25  — 3c 

The  difference  may  be  put  under  the  form  Aa—{2h  —  Sc). 
We  must  now  remark  that  it  is  the  difference  between  21) 
and  3  c  which  is  to  be  taken  from  4a. 

If  then,  we  write 4a— 25, 

we  shall  have  taken  away  too  much  by  the  units  in  3c ; 
hence,  3c  must  be  added  to  give  the  true  remainder,  which 
is 4a— 25+3c. 

To  illustrate  this    example   by  figures,  suppose    a =5, 
5z=:5,    and    c=3. 

We  shall  then  have Aa=i20 

and 25  — 3c  =10— 9   —   1 

which  may  be  written       .     4a— (25  — 3c)=:20  — 1    ==.19. 


Quest. — 29.  If  25 — 3c  is  to  be  taken  from  Aa,  what  is  proposed  to 
be  done  ?  If  you  subtract  26  from  4a,  have  you  taken  too  much  ?  How 
then  must  you  supply  the  deficiency  ? 


SUBTRACTION.  25 

Here  it  is  required  to  subtract  1  from  20.  If,  then,  we 
saotract  25  =  10,  from  4a  =  20,  it  is  plain  that  we  shall 
have  taken  too  much  by  3c  —  9,  which  must  therefore  be 
added  to  give  the  true  remainder. 

30.  Hence,  for  the  subtraction  of  algebraic  quantities, 
we  have  the  following  general 

I.  Write  the  quantity  to  be  subtracted  under  that  from  which 
it  is  to  be  taken,  placing  the  similar  terms,  if  there  are  any, 
under  each  other. 

H.  Change  the  signs  of  all  the  terms  of  the  polynomial  to 
be  subtracted,  or  conceive  them  to  be  changed,  and  then  reduce 
the  polynomial  result  to  its  simplest  form, 

EXAMPLES. 

( 1 )  |i|  ( 1 ) 

From     Qac—bab-^   c^  « ©  ^  6ac—bab-{-   c^ 

Take      3ac+3«&4-7c  ^%^       Sac~3ab  —  7c 


Rem.      3ac  —  8ab-{-   c^  — 7c.  b^^%i        Sac—Sab-^-   c^—7c, 

(2)  (3) 

From         Gax—a-^-Sb"^  6yx  —  3x^-i-5b 

Take  9ax—x-\-   b^  yx~3    -\-   a 


Rem.      —3ax^a-\-x-^2b'^,    ,  byx—3x^-\-3-{-bb  —  a, 

(4)  (5) 

From         5a3_4a25-}-   3^2^  4^5 _   cd-\-3a^ 

Take      — 2^^+ Sa^^—   %b'^c  bab—A.cd^3a^^bb'^ 

Rem.          la^—la^b^WbH.  —  ab-{-37d—5b\ 


Quest. — 30.  Give  the  rule  for  the  subtraction  of  algebraic  quantities 

3 


26  FIRST    LESSONS    IN    ALGEBRA. 

6.  From    Qam-{-y   take    Sam—^x.  Ans.  3am-{-x+i/. 

7.  From    3ax   take    3ax—7/.  Ans.  +y. 

8.  From    7a-b^~x^   take    I8a^^+x'^, 

Ans.   —1 1^252  _2a:2. 

9.  From    —7f+3m—8x   take    —6f—5m—2x+3d+8. 

Ans.   —f-i-Sm—6x~3d—S, 

10.  From    —a—5b-\-7c—d   take    4^>— c  +  2J+2^. 

^n^.   — a— 9J  +  8c— 3(^— 2^. 

11.  From  .  .    —3a-\-b-'Sc+7e—5f+3h—7x—l3y   take 
k-{-2a--9c+8e—7x+7f—7/--'3l^k. 

Ans.   — 5a+54-c— c— 12/+3A— 12y+3Z. 

12.  From   a-{-b   take    a— Z>.  ^?i^.  2b. 

13.  From   2a:— 4a-.2^>  +  5    take    8  —  5b-{-a-i-6x. 

Ans.  —ix—5a+3b—3. 

14.  From    3a-\-b-\-c—d—lO   take    c+2«— c?. 

JiW-y.  Gf+6 — 10. 

15.  From    3a+b+c—d—l0   take    5— 19  +  3«. 

^?i5-.  c— J4-9. 

16.  From   2ab+b^—ic  +  bc—b   take    3«2— c+Z*2, 

^w^.  2ab-^3a^—3c-^bc'-b, 

17.  From   a^+3i2c-|.«j2_^j^   take    b^-^-ab^-^abc. 

Ans.  a^+3b-C'^b^. 

18.  From    12a;+6«-.4i+40  take   4Z'— 3a+4a?+6J--10. 

Ans.  8x-^9a—8b--ed+50, 

19.  FroHi2a;— 3«+4^+6c— 50  take  9a  +  a:+6Z>— 6c— 40. 

^W5.  ic— 12a— 25+12c— 10. 

20.  From    6a— 45— 12c+12a?   take    2a:— 8a+45— 6c. 

Ans.  14a— 85— 6c+ 10a:. 

21.  From  8a5c— 1253a+6ca:— 7a:y  take   7ca:— o-y— IS^^a. 

Ans.  Sabc-i-b^a-^cx—dxy, 


SUBTRACTION* 


27 


31.  By  the  rule  for  subtraction,  polynomials  may  be 
subjected  to  certain  transformations. 


For  example, 
becomes  .  .  . 
In  like  manner 
becomes 
or,  again, 
Also,  .  .  . 
becomes 
Also,  .  .  . 
becomes 


6a2—  3ab  +  2b^  — 2^c, 
6a'^—{3ah  —  2b^  -\-2bc), 
7^3  _  Sa'^b—  462c +652, 
7a^'^{8a^b+  4b'^c—6b'^), 
7^3—  8a^b-(4b^c-6b^). 
8a3—  7*2  +     c    —J, 
8^3  _  (752  _    c    +d). 
9^3  _     a     +  3a2  —d, 
953 _   (^     _  3^2  _|_^), 


32.  Remark. — From  what  has  been  shown  in  addition 
and  subtraction,  we  deduce  the  following  principles. 

1st.  In  algebra,  the  words  add  and  sum  do  not  always,  as 
in  arithmetic,  convey  the  idea  of  augmentation;  for  a—b, 
which  results  from  the  addition  of  —b  to  a,  is  properly 
speaking,  a  difference  between  the  number  of  units  ex- 
pressed by  a,  and  the  number  of  units  expressed  by  b. 
Consequently,  this  result  is  numerically  less  than  a.  To 
distinguish  this  sum  from  an  arithmetical  sum,  it  is  called 
the  algebraic  sum. 

Thus,  the  polynomial  2a'^—3a'^b-^3b^c  is  an  algebraic 
sum,  so  long  as  it  is  considered  as  the  result  of  the  union 


Quest. — 31.  How  may  you  change  the  form  of  a  polynomial  ?  32.  In 
algebra  do  the  words  add  and  sum  convey  the  same  idea  as  in  arithme- 
tic ?  What  is  the  algebraic  sum  of  9  and  — 4?  Of  8  and  — 2? 
May  an  algebraic  sum  ever  be  negative  ?  What  is  the  sum  of  4  and 
—  8 1  Do  the  words  subtraction  and  difference  in  algebra  always  con- 
vey the  idea  of  diminution  ?  What  is  the  algebraic  difference  between 
8  and  — 4?     Between  a  and  — hi 


28'  FIRST  LESSONS  IN  ALGEBRA. 

of  the  monomials  2c^^  —^oF'b,  +3^2^,  with  their  respec- 
tive signs  ;  and,  in  its  proper  acceptation,  it  is  the  arithmeti- 
cal difference  between  the  sum  of  the  units  contained  in  the 
additive  terms,  and  the  sum  of  the  units  contained  in  the 
subtr active  terms. 

It  follows  from  this,  that  an  algebraic  sum  may,  in  the 
numerical  applications,  be  reduced  to  a  negative  number,  or 
a  number  affected  with  the  sign  — . 

2nd.  The  words  subtraction  and  difference  do  not  always 
convey  the  idea  of  diminution ;  for,  the  numerical  difference 
between  -\-a  and  — h  being  a-\-h,  exceeds  a.  This  result 
is  an  algebraic  diff^erence,  and  can  be  put  under  the  form  of 

a—{ — b)z=za-{-h. 


MULTIPLICATION. 

33.  If  a  man  earns  a  dollars  in  one  day,  how  much 
will  he  earn  in  6  days  ?  Here  it  is  simply  required  to  re- 
peat the  number  a,  6  times,  which  gives  Qa  for  the  amount 
earned. 

1.  What  will  ten  yards  of  cloth  cost  at  c  dollars  per  yard  1 

Ans.   10c  dollars. 

2.  What  will  d  hats  cost  at  9  dollars  per  hat  ? 

Ans.  9d  dollars. 

3.  What  will  b  cravats  cost  at  40  cents  each  ? 

Ans.  405  cents. 

4.  What  will  b  pair  of  gloves  cost  at  a  cents  a  pair  ? 


Quest. — 33.  What  is  the  object  of  multiplication  in  algebra  1  If  a 
man  earns  a  dollars  in  one  day,  how  much  will  he  earn  in  4  days  1  In 
6  days  1    In  6  days  1 


MULTIPLICATION.  89 

Here  it  is  plain  that  the  cost  will  be  found  by  repeating  h 
as  many  times  as  there  are  units  in  a  :  Hence,  the  cost  is 
ah  cents.  Ans.  ah  cents. 

Note. — If  we  suppose  «— 6,  c=4,  and  c?z=3,  w^hat  would 
be  the  numerical  values  of  the  above  answers  ? 

5.  If  a  man's  income  is  2a  dollars  a  week,  how  much 
will  it  be  in  4^  weeks.  Here  we  must  repeat  3a  dollars  as 
many  times  as  there  are  units  in  Ah  weeks  ;  hence,  the  pro- 
duct is  equal  to 

2axAh  =  l2ah, 

If  we  suppose  a=z4:  and  5  =  3  the  product  will  be  equal 
to  144. 

34.  Remark.— It  is  plain  that  the  product  I2ah  will  not 
be  altered  by  changing  the  arrangement  of  the  factors  ;  that 
is,  \2ah  is  the  same  as  «5xl2,  or  as  hax\2,  or  as 
ax  12x5  (See  Arithmetic,  ^  22). 

35.  Let  us  now^  multiply  3a252  ^y  2a%,  which  may  be 
placed  under  the  form 

3a252  X  2a%  =  3  X  2aaaahhh  ; 

in  which  a  is  a  factor  four  times,  and  h  a  factor  three  times: 
hence  (Art.  13). 

3^252  X  2a25  =  3  X  2aaaahhh=z6a'^h^, 

in  which,  we  multiple/  the  co-efficients  together  and  add  the 
exponents  of  the  like  letters. 


Quest. — 34.  Will  a  product  be  altered  by  changing  the  arrangement 
of  the  factors  1  Is  Sab  the  same  as  Sba  1  Is  it  the  same  as  aXSbl 
As  5  X  3a  T  35.  In  multiplying  monomials  what  do  you  do  with  the  co- 
efficients 1  What  do  you  do  with  the  exponents  of  the  common  letters  ? 
If  a  letter  is  found  in  one  factor  and  not  in  the  other,  what  do  yo^i  do  7 

3* 


80 


FIRST  LESSONS  IN  ALGEBRA. 


Hence,  for  tlie  multiplication  of  monomials,  we  have  the 
following 

RULE. 

I    Multiply  the  co-effccients  together. 

II.  Write  after  this  product  all  the  letters  which  are  com- 
mon to  the  multiplicand  and  multiplier,  affecting  each  letter 
with  an  exponent  equal  to  the  sum  of  the  two  exponents  with 
lohich  this  letter  is  affected  in  the  two  factors. 

III.  If  a  letter  enters  into  but  one  of  the  factors,  write  it  in 
the  product  with  the  exponent  with  which  it  is  affected  in  the 

factor. 

EXAMPLES. 

1.  Sa^bc"^  X  Habd"^  =  56a^'^c^^d\ 

2.  21  a^^cd  X  8abc^  =z  1 6Sa^b^cM. 

3.  4abcx7df  =  2Sabcdf 

(4)  (5) 

Multiply  3a^  12a^-x 

by  2a^  \2x'^y 


(6; 

Qxy  z 
ay'^z 


Ga'^b'^ 


144a'^x^y  Saxy^z^, 


(7) 

(P'xy 

2xy'^ 

2a^x'^y^ 


(8) 

3«Z»2c3 

^a^b^c 


(9) 

87  ax'^y 
3^*y3 


10.  Multiply    5aWx^  by  6c V. 

11.  Multiply  lOa'^b^c^  by  7acd. 

12.  Multiply  9a%xy  by  9a^xy. 

13.  Multiply  36a%'^c^d^  by  20abVd\ 

14.  Multiply  27 axyz  by  9a%'^c'^d'^xyz. 

Ans. 

15.  Multiply  IZc^h^c  by  Sahxy. 


261  ab^xY- 

Ans.  30^352^5^8^ 

Ans.  70a^¥c^d. 

Ans.  8la%^x^y'^. 

Ans.  720a%^c^d\ 

Ans    \04a^l^cxy 


MULTIPLICATION.  31 

16.  Multiply  20a^h^cd  by  12a^x^y.      Ans.  2\0d^¥cdx^y. 

17.  Multiply  lAa^hH^y  by  20a?c^xhj. 

Ans.  2S0a'¥cH^xh/. 

18.  Multiply  Sa^Z^y  by  la^hxi/.  Ans.  bQa^h^xy^. 

19.  Multiply  Ibaxyz  by  ba^hcdx^y'^ . 

Ans.  37da%cdx^y^z, 

20.  Multiply  51«2y2^2  by  9a?hc^x^y. 

Ans.  459a'^bc^x'^y^, 

36.  We  will  now  proceed  to  the  multiplication  of  poly- 
nomials. Take  the  two  polynomials  a-\-b-{-c,  and  d-\-fj 
composed  entirely  of  additive  terms ;  the  product  may  be 
presented  under  the  form  {a-{-b-]-c)  {d+j).  It  is  now  re- 
quired to  take  the  multiplicand  as  many  times  as  there  are 
units  in  d  s,ndf. 

Multiplicand a-{-b-{-c 

Multiplier d-^f 

taken  d  times ad-\-bd-\-cd 

taken/times +af-\-bf-j-cf 

entire  product      ....     ad-^-bd-^-cd-^-af-j-bf-i-cf. 

Therefore,  in  order  to  multiply  together  two  polynomials 
composed  entirely  of  additive  terms  : 

Multiply  successively  each  term  of  the  multiplicand  by  each 
term  of  the  multiplier,  and  add  together  all  the  products, 

EXAMPLES. 

1.  Multiply 3fl2+   4:ab  +  b^ 

by 2a  +  5b 

6a3+   Sa%'\-2ab'^ 
The  product,  after  reducing,  -{'\ba%-\-20ab'^+bb^ 

becomes       ....  Qa^-^23a:^b  +  22ab'^-\-bbK 


Quest. — 36.    How  do  you  multiply  two  polynomials  composed  of 
additive  terms  1 


33  FIRST    LESSONS    IN    ALGEBRA. 

(2)  (3) 

x^+y^  a;^+   ocy^    +7  ax 

X  +y  CLX  -\-bax 


x^+xy'^  ax^+   ax'^y^-\-ld^x'^ 

a?34-a?y2^_-jn2y_|_y3  6aa;^+6aa?y-j-42a^a?2. 

4.  Multiply   x'^-\'2ax+d^   by   x-\-a. 

Ans.  x^-\-Zax'^+^a?x+a^ 

5.  Multiply   x^+y^   hj   x+y. 

Ans.  x^-\-xy'^-\-x'^y-\-y^ 

6.  Multiply    3«52+66fV   by    Sai^+Sa^c^. 

Ans.  9a2J4_(.  27^3^,2^2 _|_i8aM 

7.  Multiply    a25^+c2(^   by   a+h. 

Ans.  a?b'^+ac^d+a%'^-{-hc^d, 

8.  Multiply    Zax'^+9ah'^-\'Cd^   by    Ga^cS. 

9.  Multiply   64a3a;3  +  27a2a:+9a5   by    Sa?cd. 

Ans.  bl2a^cdx^+2\Qa^cdx  +  l[2a^hcd, 

10.  Multiply   a'^+2ax-\-x'^   by   a+o:. 

11.  Multiply    a^+^aH+^ax'^-{-x^   by   «+«?. 

Ans.  a!^-\-Aa'^x+Qa?'x'^+Aax^-{-x^, 

37.  To  explain  the  most  general  case,  multiply   a—h 
by   c—d. 

The  required  product  is  equal  to  a  —h 

a—b  taken  as  many  times  as  there  c  —d 

are  units  in  c—d.     If  then  we  mul-  ac  —  bc 

tiply  by  c,  which  gives  ac  —  bc,  we  —ad+bd 

have  got  too  much  by  a—b  taken  ac—bc—ad-j-bd. 
d  times ;  that  is,  we  have   ad—db 


MULTIPLICATION.  33 

too  much.     Changing  the  signs,  and  subtracting  this  from 
the  first  product  (A.rt.  30),  we  have 

(«  —  b)  (c — d)=zac  —  be — ad-\-bd. 

Let  us  suppose  (2=10,  bz^iQ,  <:  =  5,  and  d=^\\  in  which 
case  we  find  the  product 

{a  —  b)  (c — d)=zac  —  be  —  ad-\-bd=i\Q. 

Hence,  we  have  the  following  rule  for  the  signs. 

When  two  terms  of  the  multiplieand  and  multiplier  are 
ajfected  with  the  same  sign,  the  corresponding  product  is  affect- 
ed with  the  sign  -\-  ;  and  when  they  are  affected  with  contrary 
signs,  the  product  is  affected  with  the  sign   — . 

Therefore  we  say  in  algebraic  language,  that  +  multi- 
plied by  -|-,  or  —  multiplied  by  —,  gives  +5  —multi- 
plied by  +)  01*   +   multiplied  by   — ,  gives   — . 

Hence,  for  the  multiplication  of  polynomials  we  have  the 
following 

RULE. 

Multiply  all  the  terms  of  the  multiplicand  by  each  term  of 
the  multiplier,  observing  that  like  signs  give  plus  in  the  pro- 
duct, and  unlike  signs  minus.  Then  reduce  the  polynomial 
result  to  its  simplest  form. 

EXAMPLES- 

1.  Multiply     ....  2ax  —   ^ah 

by ^x    —     b. 

The  product Qax"^—   9abx 

becomes  after     ....  —  2a:bx-^3ab^ 

reducing 6ax^~llabx-^3ab^. 

QoEST. — 37.  What  does  +  multiplied  by  +  give?  +  multiplied 
by  — 1  —  multiplied  by  +  ?  —  multiplied  by  —  ?  Give  the  rule  for 
the  multiplication  of  polynomials  ? 


84  FIRST    LESSONS    IN    ALGEBRA. 

2.  Multiply   a*—2P   by   a  —  h. 

Ans.  a^—2aP—a^h+2h\ 

3.  Multiply   a;2--3a?— 7   by   x—2. 

Ans.  a?^— 50:^— a:-|-l4. 

4.  Multiply    3a^-^5ab-\-2b^   by   a^  —  7ab, 

Ans.  3a'^—26a^-\-37a^^—Uab\ 
6.  Multiply   b^+¥+¥   by   b^—1.  Ans.  b^—b^. 

6.  Multiply   x'^—2x^y+4xh/  —  8xy^-{-l6i/   by    a:+2y. 

JLw5.  a:^-f  32y^. 

7.  Multiply   4a;^— 2y  by   2y.  Ans,  Sx^i/  —  4i/^, 

8.  Multiply   2a?+4y   by   2a?— 4y.  Ans.  Ax'^—ldj/'^. 

9.  Multiply   x^+x^y+xj/^+y^   by   a:— y. 

10.  Multiply   x^-{-xy+y^   by   a?^— a:y+y^- 

^71^.  a:*+a?y+y*- 
IL  Multiply   2^—3ax+4x^  by   5a2_6^.p_2a;2^ 

Ans.  10a^—27a^x-^3ia'^x^-'l8ax^—8x\ 

12.  Multiply   3a?2— 2a;y+5   by   x^+2xy—3. 

Ans.  3x^ + 4x^y  —  4a?2  —  4a?2y^  + 1 6xy-- 1 5. 

13.  Multiply    3x^+2x^y^  +  3y^   by   2a;3_3a:y  +  5y3. 

.        (  6a:6  — 5a;5y2_6a:^y4_^6a:3y24. 
^^*  <  15a:3y3-9a^y+10a:2y5-M53^. 
i       14.  Multiply    8ax — 6ab—c   by   2aa;  +  aJ+c. 

Ans,   1 6fl52aj2  __  4gj2^-j.  _  6a2Z>2  +  6«ca;  —  7a5c  —  c\ 

15.  Multiply    3a2— 552+3c2   by   a^  —  b^. 

Ans.  3a*-8a252+3a2c2+5i*~35V. 

16.  3a^  —  5bd+  cf 

—  5g24-4g>^— 8c/: 

PK).red.    ~  1 5a^4- 37aHd--29a'^cf—20b'^d'^-h44bcdf~ 8c'^p. 


MULTIPLICATION.  ./         35 

38.  To  finish  with  what  has  reference  to  algebraic  mul- 
tiplication, we  will  make  known  a  few  results  of  frequent 
use  in  Algebra. 

Let  it  be  required  to  form  the  square  or  second  power  of 
the  binomial  {a-\-b).     We  have,  from  known  principles, 

{a+b)^={a+b)  (a+b)=za^+2ab+b\ 

That  is,  the  square  of  the  sum  of  two  quantities  is  equal  to 
the  square  of  the  first,  plus  twice  the  product  of  the  first  by  the 
second,  plus  the  square  of  the  second, 

1.  Form  the  square  of  2fl+35.     We  have  from  the  rule 

{2a  +Uf      =   4a2    +    \2ab    -h   9R 

2.  {bab+^acf     =z2^a%'^+   ZOa'^bc+  ^aW 

3.  (pa^+Ba%f  z=:25a^    +   SOa^b  +Ua^b\ 

4.  (6«a;+9a2a:2)2  =  36a2a;2_f.i08a3a;3+81a%*. 

^'    39.  To  form  the  square  of  a  difference  a— 5,  we  have 

(a-bf^ia-b)  {a^b)-a^-'2ab+b\ 

That  is,  the  square  of  the  difference  between  two  quantities  is 
equal  to  the  square  of  the  first,  minus  twice  the  product  of  the 
first  by  the  second,  plus  the  square  of  the  second, 
1    Form  the  square  of  2a— b.     We  have 

(2a-Z>)2=:4a2~.4a5+R 

2.  Form  the  square  of  4ac— ic.     We  have 

(4«c  -  bcf  =  1 6a2c2  -  Sabc'^ + b'^c'^, 

3.  Form  the  square  of  7a^b'^—l2ab^.     We  have 

(7^252  _i2a53)2_  49^4^,4  _i68a365+ 144^2^6. 

Quest. — 38.  What  is  the  square  of  the  sum  of  tw6  quantities  equal  to  t 
39.  What  is  the  square  of  the  difference  of  two  quantities  equal  to  1 


86  FIRST   LESSOKS    IN    ALGEBRA, 

40.  Let  it  be  required  to  multiply  «-f6  by  a—b.  We 
liave 

{a+b)x(a-b)  =  a^-P. 

Hence,  the  sum  of  two  quantities,  multiplied  by  their  differ' 
ence,  is  equal  to  the  difference  of  their  squares. 

1.  Multiply   2c+b  by   2c— b.     We  have 

(2c+b)  X  (2c— Z>)  =  4c2— Z>2.       , 

2.  Multiply    9ac-f  35c   by    9ac—3bc.     We  have 

{9ac+3bc){9ac  —  3bc)  =  Sla'^c'^  —  9b^c^. 

3.  Multiply    Sa^+7ab^   by    8a^—7ab'^.     We  have 

(8a^+7ab^){Sa^—7ab^)z=z6ia^—i9a'^b\ 

41.  It  is  sometimes  convenient  to  find  the  factors  of  a 
polynomial,  or  to  resolve  a  polynomial  into  its  factors. 
Thus,  if  we  have  the  polynomial 

ac-\-ab-\-ad, 

we  see  that   a   is  a,  common  factor  to  each  of  the  terms : 
hence,  it  may  be  placed  under  the  form 

a{c+b-\-d). 

1.  Find  the  factors  of  the  polynomial    a'^b^-\-a^d-\-a'^f 

Ans.  a'^{b'^-\-d-\-f). 

2.  Find  the  factors  of  Sa^  +  Ga^^-^b^d, 

Ans.  b{3a^-{-Ga^b  +  bd). 

3.  Find  the  factors  of  3a'^b+9a'^c-{-l8a'^xi/. 

Ans.  3^2(5 -f  3c +6a?y). 

QuERT. — 40.  What  is  the  sum  of  two  quantities  multiplied  by  their 
difference  equal  to  1 


DIVISION.  37 

4.  Find  the  factors  of  8a^cx—lSacx^+2ac^i/'^30a^c^x, 

Ans.  2ac(4cfa;—9a;2+c*y— ISa^c^ir), 

5.  Find  the  factors  of  a^+2ah+b^. 

Ans,  {a+b)x{a+h). 

6.  Find  the  factors  of  o^—bK  Ans,  {a+b)x{a—by 

7.  Find  the  factors  of  a^—Hab+b^ 

Ans.  (a--5)x(a— 5). 


S.  DIVISION. 

42.  Algebraic  division  has  the  same  object  as  arithmeti- 
cal, viz :  having  given  a  product,  and  one  of  its  factors,  to 
find  the  other  factor. 

We  will  first  consider  the  case  of  two  monomials. 

The  division  of  72a^  by   8a^   is  indicated  thus  : 

It  is  required  to  find  a  third  monomial,  which,  multiplied 
by  the  second,  will  produce  the  first.     It  is  plain  that  the 
third  monomial  is  9a^ ;  for  by  the  rules  of  multiplication 
Sa^x9a'^=72a^ 

Hence,  we  have  =9g^, 

a  result  which  is  obtained  by  dividing  the  coefficient  of  the 
dividend  by  the  coefficient  of  the  divisor,  and  subtracting  the 
exponents  of  the  like  letter. 

Quest. — 42.  What  is  the  object  of  division  iti  Algebra  1  Give  the 
rule  for  dividing  monomials  1 

4 


30  FIRST  LESSONS  IN  ALGEBRA. 

Also,  I^^^=^5a^-ih^'^c  =  5a^c, 

for,  7ah  x  da^c  =  35a^^c,  • 

Hence,  for  the  division  of  monomials  we  have  the  foUovring 

I.  Divide  the  coefficient  of  the  dividend  hy  the  coefficient  of 
the  divisor. 

II.  Write  in  the  quotient,  after  the  coefficient,  all  the  letters 
common  to  the  dividend  and  divisor,  and  affect  each  with  an 
exponent  equal  to  the  excess  of  its  exponent  in  the  dividend 
over  that  in  the  divisor. 

III.  Annex  to  these,  those  letters  of  the  dividend,  with  their 
respective  exponents,  which  are  not  found  in  the  divisor. 

From  these  rules  we  find 

^Ba%^c'^d       ,  ,,    ,        IbOa^h^cd"^        ^  ^,,   , 

=:4a'^oca:       :; — - —   zzzocrb^cd. 

12ab^c  '        SOa^'d^ 

1.  Divide  I6x^   by    8a?.  Ans.  2x. 

2.  Divide  I5axy^   by    Say.  Ans.  5xy^. 

3.  Divide  S4aPx  by    I2b^.  Ans.  7abx. 

4.  Divide  36a^b^c^   by    9a^^c.  Ans.  4ab^c. 

5.  Divide  88^352^   by    Sa'^b.  Ans.   Uabc. 

6.  Divide  99a^b^x^   by    lla^^x^  Ans.  9ab^x. 

7.  Divide  I08x^y^z^   by   bix^z.  Ans.  2xy^z^. 

8.  Divide  64x'^y^z^   by    16x^y^z^.  Ans.  Axyz. 

9.  Divide  9QaWc^   by    \2a%c.  Ans.  Qa^¥c^, 

10.  Divide    bid^c^d^   by   27 acd.  Ans.  2a^c^d^. 

11.  Divide    ZQa'^¥d^  by   2a%H.  Ans.  \9abd\ 


DIVISION. 


39 


12.  Divide  42a^b^c^   by    7ahc.  Ans.  dahc, 

13.  Divide  64a^¥c^   by    32a*5c.  Ans,  2aPc\ 

14.  Divide  12Sa^x^i/   by    IQaxy^,  Ans,  Qa^x^y^. 

15.  Divide  U2hd^p   by   2#/.  An^.  6656^/5. 

16.  Divide  256^4^,9^8^7    by    IQa^hc^,  Ans,  IQab^c^dP, 

17.  Divide  200a^m?n^   by    50aPmn.  Ans.  4amn, 

18.  Divide  SOOx^y^z^   by    60a:y2^.  ^;^^^  bx^y^z, 

19.  Divide  27 a^b^c^   by    9a5c.  ^w^.  3«*5c. 

20.  Divide  64a^y^z^   by    32fl5y5;j7^  ^^^^^  2a2y;3'. 

21.  Divide  SSa^b^c^   by    lla^Z^^c^.  Ans.  Qa%'^c\ 

43.  It  follows  from  the  preceding  rule,  that  the  division 
of  monomials  v^^ill  be  impossible, 

1st.  When  the  coefficients  are  not  divisible  by  each  other. 

2nd.  When  the  exponent  of  the  same  letter  is  greater 
in  the  divisor  than  in  the  dividend. 

3rd.  When  the  divisor  contains  one  or  more  letters  which 
are  not  found  in  the  dividend. 

When  either  of  these  three  cases  occurs,  the  quotient  re- 
mains under  the  form  of  a  monomial  fraction ;  that  is,  a 
monomial  expression,  necessarily  containing  the  algebraic 
sign  of  division,  but  which  may  frequently  be  reduced. 

Take,  for  example,  I2a'^b^cd,  to  be  divided  by  8a^bc\ 
which  is  placed  under  the  form 

na'^b^cd  ^ 
Sa?bc^     ' 


Quest. — 43.  What  is  the  first  case  named  in  which  the  division  of 
monomials  will  not  be  exact  1  What  is  the  second  1  What  is  the  third  1 
If  either  of  these  cases  occur,  can  the  exact  division  be  made  1  Under 
what  form  will  the  quotient  then  remain  1  May  this  fraction  be  often 
reduced  to  a  simpler  form  1 


40  FIRST    LESSONS    IN    ALGEBRA. 

this  may  be  reduced  by  dividing  the  numerator  and  denomi- 
nator by  the  common  factors  4,  «^,  h,  and  c,  which  gives 


Sa%c^  2c 


44.  Hence,  for  the  reduction  of  a  monomial  fraction  we 
have  the  following 

RULE. 

I.  Suppress  the  greatest  factor  common  to  the  two  co- 
efficients. 

II.  Subtract  the  less  of  the  two  exponents  of  the  same  letter 
from  the  greater,  and  write  the  letter  affected  with  this  differ- 
ence in  that  term  of  the  fraction  corresponding  with  the  greatest 
exponent. 

III.  Write  those  letters  which  are  not  corrmion,  with  their ^ 
respective  exponents ^in  the  term  of  the  fraction  which  contains 
them. 

From  this  new  rule,  we  find, 

(1)  (2) 

ASaWcd?       4ad2  ,     37a  Pc^d        37b^c 

and 


dea^^c^de'' 3bce  '  6a^c^d^~~    da'^d    ' 

(3)  (4) 

7a'^b  I  ,         4a2Z>2  2a 

also    — ...070    =  A  .   *     and 


Ua^b^    '~  2ab  '  Qa¥  3b^ 

7bc^ 

5.  Divide   49a^^c^  by    I4a^bc\  Ans.  — — 

2a 

6.  Divide   6amn  by   3abc.  Ans.  -^ — . 

•^  be 

7.  Divide    }8c^b^mn^   by    I2a^¥cd.  Ans.  tt^-j- 

•^  2a^b^^cd 

Quest. — 44.  Give  the  rule  for  the  reduction  of  a  monomial  fraeticMx 


DIVISION.  41 


8.  Divide    l^a^h^c^d^   by    \^db^cd?m.  Arts. 

9.  Divide    12a?c^W'   by    Yla^c^ly^d,  Ans, 
10.  Divide    \^^a%^xmn   by   25a^^d,        Ans, 


4b^m 
6 


Aa^hxmn 


11.  Divide    9Qa^h^c^df  by    IbaHxy,  Ans,  — |r— — • 

12.  Divide    ^^mVfx^y'^   by    Ibam^nf,       Ans.    — — ^J^. 

127 

13.  Divide    I27d^xh/   by    16cZ^;ry,  An^.  TT^-g-^. 

45.  If  v^e  have  an  expression  of  the  form 

a  c^  a^  a^  a^ 

— ,     or    — ,     or     —,    or    — ,     or    — ,    &c^ 
a  a?  a^  a^  aP 

and  apply  the  rule  for  the  exponents,  we  shall  have 

But  since  any  quantity  divided  by  itself  is  equal  to  1,  it 
follows  that 

— =aO  =  l,     ^=02-2:^^^1,  &c, 
a  a^ 

or  finally,  if  we  designate  the  general  exponent  by  m,  we 
have 

or 
that  is,  any  power  of  which  the  exponent  is  0  is  equal  to  1. 


Quest. — 45.  What  is  afi  equal  to?     What  is  Jfi  equal  tol    What  is 
the  power  of  any  number  equal  to,  when  the  exponent  of  the  power  is  Ot 

4* 


42  FIRST    LESSONS    IN    ALGEBRA. 

2.  Divide    Ba^b^c^d  by   2a^b^d. 


"ZaWd 


=  3a2 -2^2-2^4^ -i^3e*. 


3.  Divide    8a*5V(?5   by   \a%'^d^d^.  Ans.  2a^, 

4.  Divide  I6a%^d^  by    S^ejsj,  Ans.  2d^. 

5.  Divide    32m^n^x'^y^   by   47n3w3a!:^.  j1»^.  8a?y, 
e.  Divide    dda'^b^d^c^   by   2U^¥d^cK  Ans.  4bd^. 

SIGNS  IN  DIVISION. 

46.  The  object  of  division,  is  to  find  a  tliird  quantity 
called  the  quotient,  which,  multiplied  by  the  divisor,  shall 
produce  the  dividend. 

Since,  in  multiplication,  the  product  of  two  terms  having 
the  same  sign  is  affected  with  the  sign  + ,  and  the  product 
of  two  terms  having  contrary  signs  is  affected  with  the 
sign  — ,  we  may  conclude, 

1st.  That  when  the  term  of  the  dividend  has  the  sign  +, 
and  that  of  the  divisor  the  sign  of  +  y  the  term  of  the  quo- 
tient must  have  the  sign  + . 

2nd.  When  the  term  of  the  dividend  has  the  sign  -f ,  and 
that  of  the  divisor  the  sign  —,  the  term  of  the  quotient 
must  have  the  sign  — ,  because  it  is  only  the  sign  — , 
which,  multiplied  with  the  sign  —,  can  produce  the  sign  + 
of  the  dividend. 


Quest.— 46.  What  will  the  quotient,  multiplied  by  the  divisor,  be 
equal  to  ?  If  the  multiplicand  and  multiplier  have  like  signs,  what  will 
be  the  sign  of  the  product?  If  they  have  contrary  signs,  what  will  be 
the  sign  of  the  product  ?  When  the  term  of  the  dividend  and  the  term 
of  the  divisor  have  the  same  sign,  what  will  be  the  sign  of  the  quotient  ? 
When  they  have  diiferent  signs,  what  will  be  the  sign  of  the  quotient  ? 


DIVISION. 


43 


3rd.  When  the  term  of  the  dividend  has  the  sign  — ,  and 
that  of  the  divisor  the  sign  + ,  the  quotient  must  have  the 
sign  — .     Again  we  say  for  brevity,  that, 

+   divided  by  +?  and  —  divided  by  — ,  give  +  ; 
—  divided  by  +,  and  -f   divided  by  — ,  give  — . 


EXAMPLES. 


1.  Divide  iax   by  —2a.  Ans.  —2x, 

Here  it  is  plain  that  the  answer  must  be  —2x\  for, 
—2a  X  —  2a?=  +  4«a?,    the  divisor 


2.  Divide 

3.  Divide 

4.  Divide 

5.  Divide 

6.  Divide 

7.  Divide 

8.  Divide 

9.  Divide 

10.  Divide 

11.  Divide 

12.  Divide 

13.  Divide 

14.  Divide 

15.  Divide 

16.  Divide 

17.  Divide 

18.  Divide 


36a3a;2   by   —I2a^x, 
-^5Sa-b^c'^(P   by  29a^^c. 

—  84«4Z>5j3   by   —42a^b'^d. 
64c^d^x^   by    IGc^dx. 

—  885Vy6   by    '-24Pcdx^, 

77aY2^   by    -ll«y^^. 
84a'^b^c'^d  by    '-42a'^b'^c^d. 

—QOd?¥c^d   by  —I2a%''c^d'^, 

—  88a%'c^   by   Sa^b^c^, 
I6x^   by    —8a?. 

—  15a2a?y3   by    2ay. 

—  Siab^x   by    —  12R 

—  96a^^c^   by    I2a^c. 

—  lUa^^c^d^  by  —SGa^b^c^d, 
256a^c^x^   by    — 16a^cx'^. 
—SOOa^b^c^x^  by  SOa'^bh'^x, 
dOOa^^c^  by  -lOOa^JV. 


Ans.  +5- 


Ans.   — Sax, 

Ans.   —2abcd^. 

Ans.  2a%W. 

Ans.       Ad'^x'^. 

Ans     +H^' 
^''''    +   2cd  ' 

Ans.   —7. 

Ans.   —2. 

1_ 

abed' 

Ans.  —Wab. 

Ans.   — 2x. 

Ans.   —baxy'^. 

Ans.  labx. 

Ans.   — Sabc^. 

Ans.  Aa^b^cd"^. 

Ans.   — 16abcx. 

Ans.   — lOabcx. 

Ans.   —  5abc^. 


44  FIRST    LESSONS    IN    ALGEBRA. 

19.  Divide  —6ia^b^c''  by  '~Sa^b''c^.  Ans.  Sabc. 

20.  Divide  +96^5^,4^9  by  —24M^d,        Ans.   —4ab^d\ 

21.  Divide  72a^Pd^  by  —Sa^b'^d,  Ans.  —9abd^. 


I 


a — X 
Quotieiit. 


Division  of  Polynomials, 

FIRST    EXAMPLE. 

47.  Divide     a'^~2ax-{-x'^     by     a—x. 

It  is  found  most  convenient,  Dividend.     Divisor. 

in  division  in  algebra,  to  place  a^  —  2ax-{-x'^ 

the  divisor  m  the  right  of  the  a^ —  ax 
dividend,    and   the    quotient    di-  —   ax-^-x"^ 

rectly  under  the  divisor.  —  ax-j-x"^ 

We  first  divide  the  term  a^  of  the  dividend  by  the  term  a 
of  the  divisor  :  the  partial  quotient  is  a,  which  we  place 
under  the  divisor.  We  then  multiply  the  divisor  by  a,  and 
subtract  the  product  a^—ax  from  the  dividend,  and  to  the 
remainder  bring  down  x'^.  We  then  divide  the  first  term  of 
the  remainder,  — ax  by  a,  the  quotient  is  — x.  We  then 
multiply  the  divisor  by  —x,  and,  subtracting  as  before,  we 
find  nothing  remains.     Hence,  a—x  is  the  exact  quotient. 

In  this  example,  we  have  written  the  terms  of  the  dividend 
and  divisor  in  such  a  manner  that  the  exponents  of  the  same 
letter  shall  go  on  diminishing  from  left  to  right.  This  is 
what  is  called  arranging  the  dividend  and  divisor  with 
reference  to  a  certain  letter.  By  this  preparation,  the  first 
term  on  the  left  of  the  dividend,  and  the  first  on  the  left  of 
the  divisor,  are  always  the  two  which  must  be  divided  by 
each  other  in  order  to  obtain  a  term  of  the  quotient. 

Quest. — 47.  What  do  you  understand  by  arranging  a  polynomial  with 
reference  to  a  particular  letter  ? 


DIVISION.  45 

48.  Hence,  for  the  division  of  polynomials  we  have  the 
following 

RULE. 

I.  Arrange  the  dividend  and  divisor  with  reference  to. a  cer- 
tain letter,  and  then  divide  the  first  term  on  the  left  of  the 
dividend  hy  the  first  term  on  the  left  of  the  divisor,  the  result 
is  the  first  term  of  the  quotient ;  multiply  the  divisor  hy  this 
term,  and  subtract  the  product  from  the  dividend. 

II.  Then  divide  the  first  term  of  the  remainder  hy  the  first 
term  of  the  divisor,  which  gives  the  second  term  of  the  quotient ; 
multiply  the  divisor  hy  the  second  term,  and  suhtract  the  pro* 
duct  from  the  result  of  the  first  operation.  Continue  the  same 
process  until  you  ohtain  0  for  a  remainder ;  in  which  case  the 
division  is  said  to  he  exact. 

SECOND    EXAMPLE. 

Let  it  be  required  to  divide 

b\a%'^+\Oa^—ABa%  —  \^h^+iah^   by    4ah—5a'^+3h\ 
We  here  arrange  with  reference  to   a. 

Dividend.  Divisor. 


l0a^—48a^-\-5la'^h^-\-   4ah^  —  l5h\ 

-\-lOa^—   Sa^~   ^aW 

— 40a3Z>+57«2Z,2_^   4a63— IS^^'t 
— 40a3Z>+32a2^>2^24«53 


—  5a2-|-4a64-352 
— 2a2+8a5— 562 

Quotient. 


25a'^h^~20ah^  —  l5h^ 
25a2J2_20«^»3_i5^,4 


Quest. — 48.  Give  the  general  rule  for  the  division  of  pol)momials  1 
If  the  first  term  of  the  arranged  dividend  is  not  divisible  by  the  first  term 
of  the  arranged  divisor,  is  the  exact  division  possible  1  If  the  first  term 
of  any  partial  dividend  is  not  divisible  by  the  first  term  of  the  divisor,  is 
the  exact  division  possible  1 


46  FIRST  LESSONS  IN  ALGEBRA. 

Remark. — When  the  first  term  of  the  arranged  dividend 
is  not  exactly  divisible  by  that  of  the  arranged  divisor,  the 
complete  division  is  impossible  ;  that  is  to  say,  there  is  not 
a  polynomial  which,  multiplied  by  the  divisor,  will  produce 
the  dividend.  And  in  general,  we  shall  find  that  a  division 
is  impossible,  when  the  first  term  of  one  of  the  partial 
dividends  is  not  divisible  by  the  first  term  of  the  divisor. 

GENERAL   EXAMPLES. 

1.  Divide  ISac^   by    9a?.  Ans.  2x. 

2.  Divide  10a?y   by    —bx^y.  Ans.  —2y, 

3.  Divide  —^ax^y'^   by    ^x'^y.  Ans.  —ay. 

4.  Divide  —Sx'^   by    — 2a?.  Ans.   +4a?. 

5.  Divide  \0ah-{-l5ac   by    5a.  Ans.  25+ 3c. 

6.  Divide  Wax—bAx   by   6a?.  Ans.  5«— 9. 

7.  Divide  \Ox'^y—\by'^—^y  by  5y.     Ans.  2x'^-3y  —  l, 

8.  Divide  l2a-\-3ax—18ax'^   by  3a.     Ans.  4:-\-x—6x^. 

9.  Divide  6ax'^-{-9a'^x-{-a'^x^  by  ax.     Ans.  6x-\-9a-\-ax. 

10.  Divide    a'^-\-2ax+x^   by   a+x.  Ans.  a-\-x, 

11.  Divide    a^ — 3a^y-\-3ay'^—y^   by   a—y. 

Ans.  a^ — 2ay-{-y^. 

12.  Divide    2A.a%  —  l2aHh'^  —  Qah   by    —Qah. 

Ans.   — 4«+2a2cZ>+l. 

13.  Divide  6a?^— 96  by  3a;— 6.     Ans.  2a;3+4a:2+8a;+16. 

14.  Divide  .  .  .  a^~^a'^x+\0a^x'^—\0aV-{-5ax^—x^ 
by     a^ — 2ax-\-x'^.  Ans.  a^  —  3(i^x-^3ax^ — a?^. 

15.  Divide     48a:3— 76aa:2  — 64a2a:+105a3     by     2x  —  3a. 

Ans.  2ix^—2ax  —  35a^, 


DIVISION*  4t 

16.  Divide  y^  —  ZyV+^yH^—x^hyy^  —  ^y'^x+Zyx'^—x^.    ^ 

Ans,  y^+Sy^aj+Syic^+a?^. 

17.  Divide  Q4.a'^h^ —2ba?h^   by    SaW-{-bah\ 

Ans.  Sa?b'^  —  bah^. 

18.  Divide    Qa^-\-2^a%+22ah'^+b¥  by    ^a?-\-Aah-^h'^. 

Ans.  2a+5b, 

19.  Divide    6ax^+6ax'^y^+42a^x'^   by  ax+5ax. 

Ans.  x^-i-xy^-{-7ax. 

20.  Divide    .    .   - 1 5a^-{- 37  a^-bd-29a'^cf—20bhP+44bcdf 
~8c2/2   by    3a^-—5bd-{-cf.  Ans.    —ba^+Ud—Scf. 

21.  Divide    aj^-f  aiy-f-y*   by    x^—xy-{-y^. 

Ans.  x'^-{-xy-\-y^. 

22.  Divide    a?*— y*   by    a;— y.  u4.w^.  a;^+a;2y+a?y2  4-y3. 

23.  Divide    3a^—Sa'^b^  +  3a'^c'^+5b^  —  3b^c^   by    a2_j2^ 

^w^.   3a2_5^>2  +  3c2. 

24.  Divide   .  .  Ga:^— 5a;^y2  — Gic^y^+Gaj^y^+lSaj^yS  —  ga^^yi 
+  10a;2y5+15yS    by    3a;3+2a;2y2-j-3y2^ 

Ans.  2a;^  — 3a:2y3-f5y3. 

25.  Divide    .     —  c^+lGa^oj^— 7a6c— 4a2Z>a;— Ga^J^-j-eaca? 
by   Sax— Gab— c.  Ans,  2ax+ab  +  c,    /, 

26.  Divide    ....    3a:*+4a;3y-4a;2_4<^2y2_^16:ry-15 
by     2a:y+a:2_3.  Ans.  3x'^—2xy-\-o. 

27.  Divide     x^+32y^     by     a:+2y. 

Ans.  x^—2x^y-\-4:X^y^—Sxy^-\-l6y'^, 

28.  Divide     3a^ -26 a^-Uab^-h 37 a^^     by     2b'^-5ab 
-j_3a2  ^715.  d^—7ab* 


48  FIRST   LESSONS    IN    ALGEBRA* 


CHAPTER  IL 

Algebraic  F7^actions. 

49.  Algebraic  fractions  should  be  considered  in  the  same 
point  of  view  as  arithmetical  fractions,  such  as  |^,  W ;  that 
is,  we  must  conceive  that  the  unit  has  been  divided  into  as 
many  equal  parts  as  there  are  units  in  the  denominator,  and 
that  one  of  these  parts  is  taken  as  many  times  as  there  are 
units  in  the  numerator.  Hence,  addition,  subtraction,  mul- 
tiplication, and  division,  are  performed  according  to  the  rules 
established  for  arithmetical  fractions. 

It  will  not,  therefore,  be  necessary  to  demonstrate  those 
rules,  and  in>their  application  we  must  follow  the  procedures 
indicated  for  the  calculus  of  entire  algebraic  quantities. 

50.  Every  quantity  which  is  not  expressed  under  a 
fractional  form  is  called  an  entire  algebraic  quantity. 

51.  An  algebraic  expression,  composed  partly  of  an 
entire  quantity  and  partly  of  a  fraction,  is  called  a  mixed 
quantity. 


Quest. — 49.  How  are  algebraic  fractions  to  be  considered  1  What 
does  the  denominator  show  ?  What  does  the  numerator  show  ?  How 
then  are  the  operations  in  fractions  to  be  performed '?  60.  What  is  an 
entire  quantity  1     61.  What  is  a  mixed  quantity? 


ALGEBRAIC    FRACTIONS.  49 


To  reduce  a  fraction  to  its  simplest  terms. 

52.  The  rule  for  reducing  a  monomial  fraction  to  its  low- 
est terms  has  already  been  given  (Art.  44). 

With  respect  to  polynomial  fractions,  the  following  are 
cases  which  are  easily  reduced. 

1.  Take,  for  example,  the  expression 


a?—2ah-\-h'^' 
This  fraction  can  take  the  form 

{a+h)  {a-^h) 


{a -by 

(Art.  39  and  40).     Suppressing  the  factor   a— 5,  which 
is  common  to  the  two  terms,  we  obtain 

a-{-b 
a — b' 

2.  Again,  take  the  expression 

5a^—l0a^-\-5ab^ 
8a^—8a^ 

This  expression  can  be  decomposed  thus : 

5a{a'^—2ab-i-b^) 
8a^{a-b)        ' 

ba{a—bY 


or, 


Sa\a—by 


Quest. — 52.  How  do  you  reduce  a  fraction  to  its  simplest  terms  ^ 
5 


I 


50  FIRST   LESSONS    IN    ALGEBRA. 

Suppressing  the  common  factors  a{a—h),  the  result  is 

Sa 

Hence,  to  reduce  any  fraction  to  its  simplest  terms,  we  sup- 
press or  cancel  every  factor  common  to  the  numerator  and 
denominator. 

Note. — Find  the  factors  of  the  numerator  and  denomina- 
tor as  explained  in  (Art.  41). 

EXAMPLES. 

1.  Keduce — ^  ^  ^      to  its  simplest  terms. 

Ans. 


4a2+2ac2* 

„    „    ^  15«5c+25a9J  .        .      , 

2.  Keduce        ^^  ^  .  ^^  ^        to  its  simplest  terms. 

^a^c+baU 
Ans. 

3.  Reduce      ,  ^,-  „  „       to  its  simplest  terms. 

A  ^^ 

Ans. 


3^7 
4.  Reduce  ^  ^     ■     to  its  simplest  terms. 


^•^^■^"•^^  637^336^-                      ^""-Ti-iTia- 

6.  Reduce  -_            -   to  its  simplest  terms.     Ans.  —8. 

„    ^    ,  2A¥—3Qah^                       ^             4b-6a 

7.  Reduce  t^-— ,- — ^rTr-rr^.                  ^n^. 


48a^6*—66a5^>6*  '   Sa^-^lla^b^. 


ALGEBRAIC  FRACTIONS.  51 

^  CASE  II. 

53.  To  reduce  a  mixed  quantity  to  the  form  of  a  fraction. 

RULE. 

Multiply  the  entire  part  hy  the  denominator  of  the  fraction ; 
then  connect  this  product  with  the  terms  of  the  numerator  hy 
the  rules  for  addition,  and  under  the  result  place  the  given 
denominator, 

EXAMPLES. 

1.  Reduce   Q\  to  the  form  of  a  fraction 

43 

6x7=42:     42  +  1=43:     hence,     6j=y. 

2.  Reduce    x ^ —    to  the  form  of  a  fraction. 

X 

d^—x^      x^—{o^—x^\     2aj2 — cP-       ^ 
X = ^= .     Ans, 


ax~\~  tK 

3.  Reduce    x to  the  form  of  a  fraction. 

2a 

.         ax^x'^ 
Ans. 


2a 

2a?—7 

4.  Reduce    5  -\ •    to  the  form  of  a  fraction. 

3a; 

17a;-7 

Ans,   — . 

3a? 

5.  Reduce    1 to  the  form  of  a  fraction. 

a 

2a—x-\-l 
Ans.   . 


Quest.—- 63.  How  do  you  reduce  a  mixed  quantity  to  the  form  of  a 
fraction  ] 


6*  FIRST    LESSONS    IN    ALGEBRA. 

^ 3 

6.  Reduce     l+2a? —    to  the  form  of  a  fraction. 

10^2_}_4^_|_3 
i*>^  Ans.   - 


5x 

3c+4 

7.  Reduce    2€t+b —    to  the  form  of  a  fraction. 

iea+Sb  —  3c—4: 
Ans,  g . 

Qct  cc  ■""■  clu 

8.  Reduce  6ax-\-h to  the  form  of  a  fraction. 

4a 

18a^x-{-5ab 

Ans. . 

4a 

8  -|-  6a2  J2^4 

9.  Reduce   8+3ah ■■     ,  , —    to  the  form  of  a  frac- 

96abx^+30a'^b'^x^—8 

tion.  Ans,  ,^  -  , . 

I2abx'^ 

3  52  _  8c* 

10.  Reduce     9H r- —    to  the  form  of  a  fraction. 

a—b^ 

9«_-652-8c* 
Ans.   j^ . 

CASE   in. 
54.  To  reduce  a  fraction  to  an  entire  or  mixed  quantity. 


RULE. 

Divide  the  numerator  by  the  denominator  for  the  entire  part, 
and  place  the  remainder,  if  any,  over  the  denominator  for  the 
fractional  part. 


Quest. — 64.  How  do  you  reduce  a  fraction  to  an  entire  or  mixed 
quantity  ? 


ALGEBRAIC  FRACTIONS.  63 

EXAMPLES. 


,  T.  :.     8966 

1.  Reduce     — - —    to  an  entire  number. 
8 

8)8966( 


1120  ...  6  rem. 

Hence,  1120f=:  Ans. 

2.  Reduce     to  a  mixed  quantity. 

X 

Ans,  a- 


X 


ax — x^  .  .      , 

3.  Reduce     to  an  entire  or  mixed  quantity. 


Ans,  a—x. 


^    ,             ah— 20^  .      , 

4.  Reduce     r to  a  mixed  quantity. 


2^2 
Ans.  a — 


5.  Reduce     to  an  entire  quantity.     Ans,  a  +  x, 

a  —  x 

6.  Reduce     —    to  an  entire  quantity. 

X — y 

Ans.  x'^+xy  +  y^. 
7    Reduce to  a  mixed  quantity. 

DX 

3 

Ans.  2x—l-{-—-. 
5x 

8.  Reduce to  a  mixed  quantity. 

.   o     r.       32a2ir 
Ans.  4:X^—S-] — . 


fgl  FIRST    LESSONS    IN    ALGEBRA. 


CASE   IV. 


55.  To  reduce  fractions  having  different  denominators 
to  equivalent  fractions  having  a  common  denominator. 

RULE. 

Multiply  each  numerator  into  all  the  denominators  except  its 
own,  for  the  new  numerators,  and  all  the  denominators  together 
for  a  common  denominator. 

EXAMPLES. 

1.  Reduce  ^,  J,  and  f,  to  a  common  denominator. 

1x3x5  =  15  the  new  numerator  of  the  1st. 
7x2x5^:70  „  „  „  2nd. 

4x3x2=24  „  „  „  3rd. 

and      2  X  3  X  5=30  the  common  denominator. 

Therefore,  J^,  J§,  and  f  ^,  are  the  equivalent  fractions. 

Note. — It  is  plain  that  this  reduction  does  not  alter  the 
values  of  the  several  fractions,  since  the  numerator  and 
denominator  of  each  are  multiplied  by  the  same  number. 

2.  Reduce     —     and    —    to  equivalent  fractions  having 


aXcz=:ac  > 

hxb=h^y 


a  common  denominator. 

the  new  numerators, 
and  hxc=:bc     the  common  denominator. 

Quest. — 56.  How  do  you  reduce  fractions  to  a  common  denominator! 


ALGEBRAIC    FRACTIONS.  65 

Hence,     —     and    —     are  the  equivalent  fractions. 

3.  Reduce     —     and to  fractions  having  a  com- 

,  .  ^        ac  ^      ab-^-W' 

mon  denominator.  Ans,  7—    and    — ; . 

DC  be 

4.  Reduce     — ,     — ,     and     J,     to   fractions   having   a 

Z(X        oC 

common  denominator.        Ans,  -r — ,     --; — ,     and 


6ac  6ac  6ac 

^    T^    1  3       2a;  .  2x  .       .        ,      . 

5.  Reduce     — ,     — ,     and     a-\ ,  to  tractions  havmg 

a  common  denominator. 

9a  Sax  ,       12a'^+24x 

Arts.  -— — ,     -— — ,  and 


12a  '      12a  '  12a 

1         a  a    I  /K" 

6.  Reduce      — ,      — ,      and      — — — ,       to    fractions 

.4  o  a~\~  X 

having  a  common  denominator. 

3a+3a;       2a^^2aH  Ga^+Gic^ 

6a+6a?'     6a  +60;    '  Qa  -\-Qx  ' 

ft  Qny'  rfl  —  QQ^ 

7.  Reduce     7-^,     — - — ,     and = —     to   a   common 

Jo         DC  a 

denominator. 

5acd        \Sahdx  ,        \^a%c  —  Ibhcx"^ 

Ans.  '    ^,    ,  ,     -T-rT~r»     and 


l^hcd  '       Ibhcd  '  \bhcd 

8.  Reduce     — ,     ,     and     -,     to  a  common  de- 

^a         c  a-\-b 

nominator. 

ac^-\-c%  5a^  —  5ab^  ,  5ac^ 

Ans,  -—z —  ,     -— -r —  ,     and 


ba^c-\'babc  '      baH+babc  *  ba^c+babc 


66  FIRST    LESSONS    IN   ALGEBRA. 

CASE    V.- 

56.  To  add  fractional  quantities  together, 

RULE. 

Reduce  the  fractions,  if  necessary,  to  a  common  denomina- 
tor;  then  add  the  numerators  together,  and  place  their  sum 
over  the  common  denominator. 

EXAMPLES. 

1.  Add  f,  f,  and  |  together. 

By  reducing  to  a  common  denominator,  we  have 

6x3x5=90  1st  numerator. 

4x2x5  =  40  2nd  numerator. 

2x3x2  =  12  3rd  numerator. 

2x3x5  =  30  the  denominator. 

Hence,  the  fractions  become 

90     40     12      142 


30  '  30  '  30~"   30    ' 
which,  by  reducing  to  the  lowest  terms  become  4^. 

2.  Find  the  sum  of    -^,     -r-,     and     -pr.. 
0        d  j 

Here      axdxf—adf\ 

cxhx  f=cbf  > the  new  numerators. 

ex  bxd=:ebd  5 
And       bxd xf=  bdf     the  common  denominator. 

adf        cbf    .    ebd        adf-\-cbf-\-ebd 


Quest. — 56.  How  do  you  add  fractions. 


ALGEBRAIC    FRACTIONS.  S? 


3.  To     cf r—     add     b-\ 


Ans.  a+o-\ 


he 
4.  Add    — ,     —    and   —    together.  Ans,  x+ 


5.  Add    — -—    and    —    together.  Ans.  — — . 

6.  Add  x-\ — — ■  to  3x-\ — .         Ans,  Ax-\ — — . 

5iK  X   I    Of 

7.  It  is  required  to  add   4x,  —- — ,    and    together. 

Ans,  4a: +" 


2ax 

_      _  I 

__,     _,   _    __ 


8.  It  is  required  to  add     — ,    — ,  and   — - —  together. 

49a;+12 


Ans.  2x+- 


60 


9.  It  is  required  to  add   4a?,  — ,  and  2+—  together. 

44a:+90 

Ans,  4x-\-' — . 

45 

2x  Sx 

10.  It  is  required  to  add   3x+—  and  x — --  together. 

o  y 

Ans.  3x-\- 

11.  Required  the  sum  of    ac—— -     and     1 — -, 

oa  a 

Sa^cd—ebd-j-Sad—Sac 
'^'*^-  Fad 


45 


58  FIRST  LESSONS  IN  ALGEBRA. 


57.  To  subtract  one  fractional  quantity  from  another. 

RULE. 

I.  Reduce  the  fractions  to  a  common  denominator, 

II.  Subtract  the  numerator  of  the  fraction  to  he  subtracted 
from  the  numerator  of  the  other  fraction,  and  place  the  dif- 
ference over  the  common  denominator. 

EXAMPLES. 

3  2 

1.  What  is  the  difference  between    ---  and  — . 

7  8 

Z^^  2  _24     14_10___5_ 
Y""T"~56~56~56~28*      ^^* 


2.  Find  the  difference  of  the  fractions  —-7--  and  — . 

26  6c 

TJ^^z^         (^-"  «)x3c=:3ca;— 3ac  )     , 

•n^re,     ^       .   s     «,      .   ,     ^,    J-   the  numerators 

And,  25  X  3c = 65c     the  common  denominator. 

_-             3ca? — 3ac      Aah  —  Shx     3ca? — 3ac — 4:ah-{-Shx 
^''"'''  —637 66^-= 633 •  ^"*- 

3.  Required  the  difference  of  -— -  and  — .     Ans. 


7  5  •    35  ' 

4.  Required  the  difference  of  5y  and  -^.  Ans.  — -^. 

o  8 

5.  Required  the  difference  of  —  and  — .  Ans.  --^^. 


Quest. — 57.  How  do  you  subtract  fractions  ? 


ALGEBRAIC    FRACTIONS. 


59 


6.  Required  the  difference  between  — 7 —  and  -j. 


Ans. 


3a;  4-^ 
7.  Required  the  difference  of  — — —  and 


dx-\-ad — he 
hd         * 

2a;+7 


Ans 


5b  8 

24x+8a—l0bx—35b 
40^  • 


8.  Required  the  difference  of  3a?+—   and    oc . 

cx-\-bx — ab 


Ans,  2x-\- 


hc 


CASE    VII. 

58.  To  multiply  fractional  quantities  together. 

RULE.  ♦ 

If  the  quantities  to  be  multiplied  are  mixed,  reduce  them  to 
a  fractional  form  ;  then  multiply  the  numerators  together  for 
a  numerator  and  the  denominators  together  for  a  denominator. 


t 


a 


EXAMPLES. 


1.  Multiply 


ori. 


We  first  reduce  the  com- 
pound fraction  to  the  sim- 
ple one  ^2?  ^^^  then  the 
mixed  number  to  the  equiva- 
lent fraction  ^^  ;  after 
which,  we  multiply  the 
numerators  and  denomina- 
tors together. 


by 


8^- 


Operation. 

,     3       3 
^^    Y=42' 


3      25      75      25 

Hence,    — x — = = — . 

'     42      3       126     42 


Ans. 


25 
42* 


60  FIRST  LESSONS  IN  ALGEBRA. 

2.  Multiply     a-\ by    -r-     Fi^^st,     cH =— ^ . 

ad  a  a 

Hence,    ....     X— r= 5 — .  Ans, 

a  a  ad 

3.  Required  the  product  of    —    and    — .        Ans.  — ^r-. 

4.  Required  the  product  of   —    and 


5  2a 

3a;  3 


Ans. 


5a 


5.  Find  the  continued  product  of    — ,     and    — =-. 

^  a'         c  2b 

Ans.  9ax. 

6.  It  is  required  to  find  the  product  of    b-\ and    — . 

.        ah-\-hx 
Ans. . 

X 
^2 J2  a:^_|_J2 

7.  Required  the  product  of   — 7 and    -j— — . 

x^-b^ 
Ans. 


b^c+bc^ 


8.  Required  the  product  of    x-\ -,    and    r- 

a  a-\-b 

.        ax" — ax-\-x'^ — 1 

^ns.  ^— — . 

a^-\-ao 

(P'  —  X^ 

a — 0?     "''      X  -{-x^  ' 
;4  — «2.-v2 


9.  Required  the  product  of    a-\ by 

Ans. 


ax + ax^ — aj2 — x' 


Quest. — 58.  How  do  you  multiply  fractions  together? 


ALGEBRAIC    FRACTIONS.  64 

CASE    VIII. 

69.  To  divide  one  fractional  quantity  by  another. 

RULE. 

Reduce  the  mixed  quantities,  if  there  are  any,  to  a  fractional 
form;  then  invert  the  terms  of  the  divisor,  and  multiply  the 
fractions  together  as  in  the  last  case, 

EXAMPLES. 

10    ,         5 

1.  Divide.    •     •     •     24    ^    "¥* 

If  the  divisor  vrere  5,  the  Operation, 

quotient  would  be  ■^^-^.     But,  ^  _&      ^ 

since  the  divisor  is  |-  of  5,  8  "~         8 

the  true  quotient  must  be  8  10  10 

times  -^1^-^,  for  the  eighth  of  24  ~    "~  120 

a  number  will  be  contained  10  80        2 

v^  Q „ 

in  the  dividend  8  times  more  120         ~  120  ""  3  * 

than  the  number  itself.     In 

this  operation  we  have  actually  multiplied  the  numerator 

of  the  dividend  by  8  and  the  denominator  by  5  ;  that  is,  we 

have  inverted  the  terms  of  the  divisor  and  multiplied  the  fractions 

together. 

X..     .,  ^         ,  / 

2.  Divide     .     .     <z— --   by   ■^. 

2c  g 

h      2ac  —  b 


2c  2c 


h        f     2aC'-h       ^      2acg^hg         . 

Hence,    a-^--^^^—- X-^^ — ^  /-      •     ^^^' 

2c      ^  2c  /  2cf 


Quest, — 69.  How  do  you  divide  one  fraction  by  another  1 
6 


62  FIRST   1ESS0N3    IN    ALGEBRA. 

3.  Let    —    be  divided  by    — .  Ans.  -^tt-. 

AcS^  Ax 

4.  Let    —jr—    be  divided  by     5a?.  Ans.  -—. 

1  35 

5.  Let  be  divided  by     -— .  Ans.  . 

o                                     3  4a? 

6.  Let be  divided  by     — .  Ans. 


a?— 1  '       2  x—V 

'  ""  5a?     ,,..,,,        2a  ,  5^'a; 


7.  Let    —    be  divided  by     — .  Ans. 


3  •'35*  *     2a 


8.  Let     -— ~r     be  divided  by     ---7-.     •  Ans.  -—r-. 

Scd  ^       4d  6c^x 


Q      T      .  ^^-^^  U      A'     'A     lU  ^^  +  ^^ 


a;2— 2^>a?  +  62  -^        a?— ^>   * 

Ans.  x-\ — -. 

10.  Divide      6a2  +  _    by    c^-^-J?. 
5        -^  2 

60a2  4.25 


J.n^. 


IL  Divide      ISc^— a:4-4-     by     a^— A 
6        -^  5 


10c2—5a;+5a* 


90^2— 55a;4-5a 


5a^b  —  b^ 


12.  Divide      20a:2-4^    by     x'^-^. 


20dcYx^--8abf 
dc^x^-dc^b-hdc^' 


EQUATIONS    OF   THE   FIRST  DEGREE.  63 


CHAPTER  III. 

Of  Equations  of  the  First  Degree, 

60.  An  Equation  is  the  expression  of  two  equal  quanti- 
ties with  the  sign  of  equality  placed  between  them.  Thus, 
a?=(z+5  is  an  equation,  in  which  x  is  equal  to  the  sum  of 
a  and  5. 

6 1 .  By  the  definition,  every  equation  is  composed  of  two 
parts,  separated  from  each  other  by  the  sign  = .  The  part 
on  the  left  of  the  sign,  is  called  the  first  member ;  and  the 
part  on  the  right,  is  called  the  second  memher.  Each  mem- 
ber may  be  composed  of  one  or  more  terms.  Thus,  in  the 
equation  x-=za+h^  x  is  the  first  member,  and  a-\-h  the 
second. 

62^  Every  equation  maybe  regarded  as  the  emmciation, 
in  algebraic  language,  of  a  particular  question.  Thus,  the 
equation  a: +3?= 30,  is  the  algebraic  enunciation  of  the 
following  question : 


Quest. — 60.  What  is  an  equation  1  61.  Of  how  many  parts  is  every 
equation  composed  1  How  are  the  parts  separated  from  each  other? 
What  is  the  part  on  the  left  called '?  What  is  the  part  on  the  right  called  \ 
May  each  member  be  composed  of  one  or  more  terms  1  In  the  equation 
a; =a+ 5,  which  is  the  first  member?  Which  the  second ?  How  many 
terms  in  the  first  member  ?     How  many  in  the  second  1 


I 


64  FIRST    LESSONS    IN    ALGEBRA. 

To  find  a  number  which  being  added  to  itself,  shall  give  a 
sum  equal  ^o  30. 

Were  it  required  to  solve  this  question,  we  should  first 
express  it  in  algebraic  language,  which  would  give  the 
equation 

By  adding  x  to  itself,  we  have 
2a;r=30. 
And  by  dividing  by  2,  we  obtain 

Hence  we  see  that  the  solution  of  a  question  by  algebra 
consists  of  two  distinct  parts. 

1st.  To  express  algebraically  the  relation  between  the  knovm 
and  unknown  quantities. 

2nd.  To  find  a  value  for  the  unknown  quantity,  in  terms  of 
those  which  are  known. 

This  latter  part  is  called  the  solution  of  the  equation. 

The  given  or  known  parts  of  a  question,  are  represented 
either  by  numbers  or  by  the  first  letters  of  the  alphabet, 
a,  b,  c,  &c.  The  unknown  or  required  parts  are  repre- 
sented by  the  final  letters,   x,  y,  z,  <fec. 

EXAMPLE. 

Find  a  number  which,  being  added  to  twice  itself,  the 
sum  shall  be  equal  to  24. 


Quest. — 62.  How  may  you  regard  every  equation  1  What  question 
does  the  equation  a;+a;=30  state  1  Of  how  many  parts  does  the  solu- 
tion of  a  question  by  algebra  consist  \  Name  them.  What  is  the  2nd 
part  called  %  By  what  are  the  known  parts  of  a  question  represented  ? 
By  what  are  the  unknown  parts  represented  ? 


EQUATIONS    OP   THE    FIRST    DEGREE.  65 

Statement, 
Let  X  represent  the  number.     We  shall  then  have 

This  is  the  statement. 

Solution, 
Having    ....     a;+2a?=24, 

we  add x+2x^ 

which  gives      .     .     .  3a;=r24 ; 

and  dividing  by  3,      .  a? =8.        \  / 

63.  An  equation  is  said  to  be  verified  when  the  answer 
found,  being  substituted  for  the  unknown  quantity,  proves 
the  two  members  of  the  equation  to  be  equal  to  each  other. 

Thus,  in  the  last  equation  we  found  a; =8.  If  we  substi- 
tute this  value  for  x  in  the  equation 

a?+2a;=:24, 

we  shall  have  8+2x8=8+16  =  24. 

which  proves  that  8  is  the  true  answer. 

64.  An  equation  involving  only  the  first  power  of  the 
imknown  quantity,  is  called  an  equation  of  the  first  degree. 

Thus,  6a:+3a?— 5  =  13, 

and  ax-\-hx-\-c:=id, 

are  equations  of  the  first  degree. 

By  considering  the  nature  of  an  equation,  we  perceive 
that  it  must  possess  the  three  following  properties  : 

QuiSST. — 63.  When  is  an  equation  said  to  be  verified  ?  64.  When 
an  equation  involves  only  the  first  power  of  the  unknown  quantity,  what 
is  it  called  ?     What  are  the  three  properties  of  every  equation  ? 

6* 


66  FIRST    LESSONS    IN    ALGEBRA. 

1st.  The  two  members  are  composed  of  quantities  of  the 
same  kind  :  that  is,  dollars  =  dollars,  pounds  =  pounds,  &c. 
2nd.  The  two  members  are  equal  to  each  other. 
3rd.  The  two  members  must  have  the  same  sign. 

65.  An  axiom  is  a  self-evident  truth.  We  may  here 
state  the  following. 

1.  If  equal  quantities  be  added  to  both  members  of  an  equa- 
tion, the  equality  of  the  members  mill  not  be  destroyed. 

2.  If  equal  quantities  be  subtracted  from  both  members  of 
an  equation,  the  equality  will  not  be  destroyed. 

3.  If  both  members  of  an  equation  be  multiplied  by  the  same 
number,  the  equality  will  not  be  destroyed. 

4.  If  both  members  of  an  equation  be  divided  by  the  same 
number^  the  equality  will  not  be  destroyed. 

Transformation  of  Equations, 

66.  The  transformation  of  an  equation  consists  in  chang- 
ing its  form  without  affecting  the  equality  of  its  members. 

The  following  transformations  are  of  continual  use  in  the 
resolution  of  equations. 

First  Transformation. 

6T.  When  some  of  the  terms  of  an  equation  are  frac- 
tional, to  reduce  the  equation  to  one  in  which  the  terms  shall 
be  entire. 

I.  Take  the  equation 

2a?      3         X 


Quest. — 65.  What  is  an  axiom  1  Name  the  four  axioms.  66.  What 
is  the  transformation  of  an  equation  ?  67.  What  is  the  first  transforma- 
tion ?  What  is  the  least  common  multiple  of  several  numbers  ?  How 
do  you  find  the  least  common  multiple  ? 


EQUATIONS    OF   THE    FIRST    DEGREE.  67 

First,  reduce  all  the  fractions  to  the  same  denominator, 
by  the  known  rule  ;  the  equation  then  becomes 


72  72     '     72 


and  since  we  can  multiply  both  members  by  the  same  num- 
ber without  destroying  the  equality,  we  will  multiply  them 
by  72,  which  is  the  same  as  suppressing  the  denominator 
72,  in  the  fractional  terms,  and  multiplying  the  entire  term 
by  72  ;  the  equation  then  becomes 

48a;-54a?+12a;=792, 
or  dividing  by  6       So?--  9x+  2rK=:132. 

But  this  last  equation  can  be  obtained  in  a  shorter  way,  by 
finding  the  least  common  multiple  of  the  denominators. 

The  least  common  multiple  of  several  numbers  is  the 
least  number  which  they  will  separately  divide  without  a 
remainder.  When  the  numbers  are  small,  it  may  at  once 
be  determined  by  inspection.  The  manner  of  finding  the 
least  common  multiple  is  fully  shown  in  Arithmetic  §  87. 

Take  for  example,  the  last  equation 

2x      3         X 

We  see  that  12  is  the  least  common  multiple  of  the  deno- 
minators, and  if  we  multiply  all  the  terms  of  the  equation 
by  12,  and  divide  by  the  denominators,  we  obtain 

8a?— 9.r+2a;  =  132. 

the  same  equation  as  before  found; 


68  FIRST  LESSONS  IN  ALGEBRA. 

68.  Hence,  to  make  the  denominators  disappear  from  an 
equation,  we  have  the  following 

RULE. 

I.  Find  the  least  common  multiple  of  all  the  denominators, 

II.  Multiply  each  of  the  entire  terms  hy  this  multiple,  and 
each  of  the  fractional  terms  hy  the  quotient  of  this  multiple 
divided  hy  the  denominator  of  the  term  thus  multiplied,  and 
omit  the  denominators  of  the  fractional  terms, 

EXAMPLES. 

1.  Clear  the  equation  of  -r-+-;z — 4=3  of  its  denomi- 
nators. Ans.  70- +537—140=105. 

2.  Clear  the  equation  -^+-3 — ^=S  of  its  denomi- 
nators.  Ans.  9a?-)-6a:— 2aj=432. 

:)^ 

if         X         sc         v 

3.  Clear  the  equation    —+-5 S'+To" ^^     ofitsde- 

nominators.  Ans,  18a?-|-12a;--4a?+3a:=720. 

4.  Clear  the  equation     -r-+-;z ^=4     of  its  denomi- 

nators.  Ans.  14a:+10a:— 35a?=:280. 

5.  Clear  the  equation — +— -=15     ofitsdenomi- 

4       5       6 

nators  Ans.  15a;~12a;+10a;=900. 


Quest. — 68.  Give  the  rule  for  clearing  an  equation  of  its  denomi- 
nators. 


EQUATIONS    OF    THE    FIRST    DEGREE.  69 


<v>  <y»  /vi  /JJ 

6.  Clear  the  equation     — — —+—+—-=12    of  its  de- 


nominators. Ans.  l8x  —  12x+9x  +  Sx=S64:, 

a       c 
~b~1 


a       c       ^ 
7.  Clear  the  equation     — — — +/=^. 


Ans.  ad—hc-\-hdf=:hdg. 


8.  In  the  equation 


ax      2c^x      ,       Uc^x      5^3        2c^       ^. 

— 1-4«= — ~ — I 3&, 

h        ah  a^  tP'  a 

the  least  common  multiple  of  the  denominators  is   c^lP' ; 
hence  clearing  the  fractions,  we  obtain 

a^J^— 2a2Jc2a:+4a4^>2_453c2^_5a6_j_2a2j2c2_3a353, 


Second  Transformation, 

69.  When  the  two  members  of  an  equation  are  entire 
polynomials,  to  transpose  certain  terms  from  one  member  to 
the  other. 

1.  Take  for  example  the  equation 

5a:— 6  =  8  +  20?. 

If,  in  the  first  place  we  subtract  2x  from  both  members, 
the  equality  will  not  be  destroyed,  and  we  have 

5x—6—2x=:S. 

Whence  we  see  that  the  term  2x,  which  was  additive 
in  the  second  member  becomes  subtractive  in  the  first. 


Quest. — 69.  What  is  the  second  transformation?  What  do  you 
understand  by  transposing  a  term  1  Give  the  rule  for  transposing  from 
one  member  to  the  other. 


70  FIRST  LESSONS  IN  ALGEBRA. 

In  the  second  place,  if  we  add  6  to  both  members,  the 
equality  will  still  exist,    and  we  have 

5a?— 6~2a?+6  =  8+6. 

Or,  since  ~6  and  +6  destroy  each  other,  we  have 

5x-'2x=S+6, 

Hence  the  term  which  was  subtractive  in  the  first  mem- 
ber, passes  into  the  second  member  with  the  sign  of 
addition. 

2.  Again,  take  the  equation 

ax+h=zd-—cx. 

If  we  add  ex  to  both  members  and  subtract  h  from 
them,  the  equation  becomes 

ax-\-b-}-cx—b=d—'Cx-\-cx—b. 

or  reducing  ax+cx=d-^b. 

When  a  term  is  taken  from  one  member  of  an  equation 
and  placed  in  the  other,  it  is  said  to  be  transposed. 

Therefore,  for  the  transposition  of  the  terms,  we  have  the 
following 

RULK. 

An7/  term  of  an  equation  may  be  transposed  from  one  mem^ 
her  to  the  other  by  changing  its  sign, 

TO.  We  will  now  apply  the  preceding  principles  to  the 
resolution  of  equations. 

1.  Take  the  equation 

4x—3=2a;+5. 


EQUATIONS    OP   THE   FIRST   DEGREE.  71 

By  transposing  the  terms    — 3   and  2ir,  it  becomes 
4a?— 2a?=5+3. 
Or,  reducing  2a: =8. 

Q 

Dividing  by  2  af=— =4. 

Verification. 

If  now,  4  be  substituted  in  the  place  of  x  in  the  given 
equation 

4a:— 3  =  2a;4-5, 

it  becomes  4  X  4—3=2  x  4+5. 

or,  13  =  13. 

Hence,  the  value  of  x  is  verified  by  substituting  it  for  the 
unknown  quantity  in  the  given  equation. 
2.  For  a  second  example,  take  the  equation 

5a;     4a;  7       13ar 


12      3  8         6 

By  making  the  denominators  disappear,  we  have 

10a;— 32a?-312=21— 52a:, 
or,  by  transposing 

10a:-32a;+52a;=21  +  312 
by  reducing  30a; =333 

333       111 

a  result  which  may  be  verified  by  substituting  it  for  x  in  the 
given  equation. 

3.  For  a  third  example  let  us  take  the  equation 
(3a— a;)(a— ^)  +  2«a;=45(a:+a). 


72  FIRST    LESSONS    IN    ALGEBRA. 

It  is  first  necessary  to  perform  the  multiplications  indica- 
ted, in  order  to  reduce  the  two  members  to  two  polynomials, 
and  thus  be  able  to  disengage  the  unknown  quantity  x,  from 
the  known  quantities.  Having  done  that,  the  equation 
becomes, 

3a'^—ax—3ab+bx-{-2axz=z4:bx+4ab, 

or,  by  transposing 

—ax-\-bx-\-2ax—4bx  ^ziab-^Sab  —  da^f 
by  reducing  ax—Sbx  =z7ab  —  3a^. 

Or,  (Art.  41).  {a-3b)xz=:7ab-3a'^. 

Dividing  both  members  by   a— 3b   we  find 

_7ab  —  3a^ 
a — 36 

Hence,  in  order  to  resolve  an  equation  of  the  first  degree, 
we  have  the  following  general 

RULE. 

I.  If  there  are  any  denominators,  cause  them  to  disappear, 
and  perform,  in  both  members,  all  the  algebraic  operations 
indicated;  we  thus  obtain  an  equation  the  two  members  of 
which  are  entire  polynomials. 

n.  Then  transpose  all  the  term^  affected  with  the  unknown 
quantity  into  the  first  member,  and  all  the  known  terms  into 
the  second  member. 

HI.  Reduce  to  a  single  term  all  the  terms  involving  x : 
this  term  will  be  composed  of  two  factors,  one  of  which  will  be 
X,  and  the  other  all  the  multipliers  of  x,  connected  with  their 
respective  signs. 

IV.  Divide  both  members  by  the  number  or  polynomial  by 
which  the  unknown  quantity  is  multiplied. 

Quest. — 70.  What  is  the  first  step  in  resolving  an  equation  of  the 
first  degree  1     What  the  second  1    What  the  third  ?    What  the  fourth  ? 


EQUATIONS    OF    THE    FIRST    I>EGREE.  73 

EXAMPLES. 

1.  Given   3a:— 2+24  =  31    to  find   x,  Ans.  xz=:3, 

2.  Given   x-\~l8=z3x~5   to  find   x.  Ans,  a;==ll— . 

3.  Given   6— 2a;+10=20--3a?— 2   to  find   x. 

Ans,  a? =2. 

4.  Given   x-{-—x-\-—x=:ll    to  find   x.  Ans,  x=z6. 

1  R 

5.  Given     2x — — a:+l=5a:— 2  to  find  x.     Ans,xz=z-=-, 

Z  I 

s     \ 

6.  Given     3«a?+— — 3  =  5ir— a     to  find     x, 

6— 3a 
Ans.  xz 


'6a-26 


7.  Given     — - — h-^=20 —    to  find    x, 

2  3  2 


Ans.  a?=23— -. 
4 


8.  Given    — - — {-—-=4 : —    to  find     x. 


o6 
Ans.  a? =3-—. 
13 


9.  Given —+x=:-- — 3     to  find     x. 

4       2  8 


10.  Given ; 4=:f    to  find    x. 

c  d  -^ 


Ans,  x=4. 


cdf+4cd 
Ans.  x=:-—^ — -7—. 
3ad--2b€ 


Y4  FIRST    LESSONS    IN    ALGEBRA. 

11.  Given    !f^-i^=4-6    to  find    .. 

56  +  95— 7c 
Ans.  37=- 


16a 


X      x—2  .   X       13 


12.  Given    — —^~=.—    ta  find     x. 

Ans.  a;  =  10. 

X         X         X         X         r     J.     n     ^ 

13.  Given r+ ^=/    ^"^  ^^^     ^• 

a        o        c        a 

abcdf 
■^^^'  ^^J^d-acd-^ahd-ahc' 

Note. — ^What  is  the  numerical  value  of  x,  when  a— I, 
5=2,  c=3,  d=\,  5  =  5,  and/=6. 

14.  Given    ^-.^-^=-12if    to  find    .-. 

Ans.  a?=14. 

15.  Given     x-^^+^^=«+l     to  find     a.. 

An5.  07  =  6. 

16.  Given    x+^+~^='ioo-AZ    to  find     x. 

4       5        D 

^715.  a;=60. 

17.  Given     2ar — = — -^ —    ^o  find     x. 


Ans.  a?=3. 


18    Given     2x-\ — ^——x+a    to  find     a;. 


^a+d 
Ans.  x=z- 


6  +  5 


ax—b    ,   a      bx       bx—a  -    , 

19.  Given     — p-+~=y — -    to  find    x. 

35 
Ans.  x=- 


'3a-2b' 


EQUATIONS    OF    THE    FIRST    DEGREE.  76 

80.  Find  the  value  of  x  in  the  equation 

{a+h){x—h)  4ab—b^  a^—hx 

7 3a  — 7-7 2x-\ . 

a — o  a-jro  o 


Ans,  dcz=- 


2b{2a^+ab—b^) 


Of  Questions  producing  Equations  of  the  First  Degree 
involving  but  a  single  unknoivn  quantity. 

71.  It  has  already  been  observed  (Art.  62),  that  the 
solution  of  a  question  by  algebra  consists  of  two  distinct 
parts: 

1  St.  To  express  the  conditions  of  the  question  algebrai- 
cally ;  and 

2-d.  To  disengage  the  unknown  from  the  known  quantities. 

We  have  already  explained  the  manner  of  finding  the 
value  of  the  unknown  quantity,  after  the  question  has  been 
stated ;  and  it  only  remains  to  point  out  the  best  methods 
of  putting  a  question  in  the  language  of  algebra. 

This  part  of  the  algebraic  solution  of  a  question  cannot, 
like  the  second,  be  subjected  to  any  well  defined  rule. 
Sometimes  the  enunciation  of  the  question  furnishes  the 
equation  immediately  ;  and  sometimes  it  is  necessary  to  dis- 
cover, from  the  enunciation,  new  conditions  from  which  an 
equation  may  be  formed. 

Quest. — 71.  Into  how  many  parts  is  the  resolution  of  a  question  in 
algebra  divided  1  What  is  the  first  step  1  What  the  second  1  Which 
part  has  already  been  explained  ?  Which  part  is  now  to  be  considered  ? 
Can  this  part  be  subjected  to  exact  rules  ?  Give  the  general  rule  for 
stating  a  question 


76  FIRST    LESSONS    IN    ALGEBRA. 

Ill  almost  all  cases,  however,  we  are  enabled  to  discover 
the  equation  by  applying  the  following 

RULE. 

Consider  the  'problem  solved,  and  then  indicate,  hij  means 
cf  algebraic  signs,  upon  the  known  and  unknown  quantities, 
the  same  operations  which  it  would  he  necessary  to  perform,  in 
order  to  verify  the  unknown  quantity,  had  it  been  known. 

QUESTION'S. 

1 .  To  find  a  number  to  which  if  5  be  added,  the  sum  will 
be  equal  to  9. 

Denote  the  number  by     x. 
Then  by  the  conditions 

x+b-9. 

This  is  the  statement  of  the  question. 
To  find  the  value  of  x,  we  transpose  5  to  the  second 
member,  which  gives 

a;  =  9  — 5  =  4. 

Verification, 
4  +  5  =  9. 

2.  Find  a  number  such,  that  the  sum  of  one  half,  one 
third,  and  one  fourth  of  it,  augmented  by  45,  shall  be  equal 
to  448. 

Let  the  required  number  be  denoted  by     x. 

Then,  one  half  of  it  will  be  denoted  by    — . 
one  third  „  „  by    -— . 


one  fourth         ,  „  by    — . 


3 

X 


EQUATIONS    OF    THE    FIRST   BEGREB  T7 


And  by  the  conditions, 

This  is  the  statement  of  the  question. 
To  find  the  value  of  x,  subtract  45  from  both  members : 
this  gives 

By  clearing  the  terms  of  their  denominators,  we  obtain 

6x-j-4x-i-3x=iS36, 

or  13a;  =  4836. 

IT  4836     ^^„ 

Hence,  a;=— — =372. 

Verijlcalion. 

372       372       ^72 

-7— +-t;-+—— +45  =  186  +  124+93+45=448. 

<*  o  4 

3.  What  number  is  that  whose  third  part  exceeds  its 
fourth  by  16, 

Let  the  required  number  be  represented  by  x.     Then, 

---0;=     the  third  part, 

-— a;=     the  fourth  part. 
And  by  the  question 

This  is  the  statement.     To  find  the  value  of  x,  we  clear 
the  terms  of  the  denominators,  which  gives 

4a?— 3a?=192. 
and  a;  =192. 

7* 


1 


78  FIRST  LESSONS  IN  ALGEBRA. 

Verification, 
192        192 


z=z64~48=zl6. 


3  4 

4.  Divide  $1000  between  A,  B  and  C,  so  that  A  shall 
have  $72  more  than  B,  and  C  $100  more  than  A. 

Let     '  x=  B's  share  of  the  $1000. 

Then  x-\-  72=     A's  share, 

and  07+172=     C's  share, 

their  sum  is  3a;+ 244  =  $1000. 

This  is  the  statement. 

By  transposing  244  we  have 

3a;=1000-244=756 

756 
and  0;=-— — =252=  B's  share. 

Hence,    x+    72  =252+   72  =  $324=     A's  share. 
And         a;+ 172  =252+172  =  $424=     C's  share. 

Verification. 
252  +  324  +  424  =  1000. 

5.  Out  of  a  cask  of  wine  which  had  leaked  away  a  third 
part,  21  gallons  were  afterwards  drawn,  and  the  cask  being 
then  guaged,  appeared  to  be  half  full :  how  much  did  it 
hold? 

Suppose  the  cask  to  have  held     x     gallons. 

Then,         —    what  leaked  away. 

o 

And  -^+  21=     what  had  leaked  and  been  drawn. 

o 

X  1 

Hence,       ---+  21=-— a:    by  the  question, 
o  2 

This  is  the  statement. 


EQUATIONS    OF    THE    FIRST    DEGREE.  79 

To  iiiid  X,  we  have 

2x-\-l26=i3^, 
and  —  X  =:  — 126, 

or  X  =:      126, 

by  changing  the  signs  of  both  members,  which  does  not 
destroy  their  equality. 


Ver  if  cation. 
^^'^-+21=42+21=63=i^ 


3 

6.  A  llsh  was  caught  whose  tail  weighed  9lb.,  his  head 
weighed  as  much  as  his  tail  and  half  his  body,  and  his  body 
weighed  as  much  as  his  head  and  tail  together ;  what  was 
the  weight  of  the  fish  ? 

Let  2x=z    the  weight  of  the  body. 

Then,  9  +  a:=:   weight  of  the  head  ; 

and  since  the  body  weighed  as  much  as  both  head  and  tail, 

2a:=:9  +  9  +  a:, 

which  is  the  statement.     Then, 

2a;— a:=18     and     a?=18. 

Verification. 

2x=z36lb.=z  weight  of  the  body. 

9-{-xz=::27lb.z=:  weight  of  the  head. 

dlb.z:^  weight  of  the  tail. 

Hence,  72lb.  =  weight  of  the  fish. 


80  FIRST  LESSONS  IN  ALGEBRA. 

7.  The  sum  of  two  numbers  is  67  and  their  difference 
1 9  :  what  are  the  two  numbers  ? 

Let  xzzz    the  least  number. 

Then,  x-\-19=z    the  greater. 

and  by  the  conditions  of  the  question 

2a;+19  =  67. 

This  is  the  statement. 

To  find  X,  we  first  transpose  19,  which  gives. 

2a;  =  67-19  =  48; 

48 
hence,         x=z—z=24,     and     a?+ 19=43. 

Verification, 
43+24=67,     and     43-24=19. 

Another  Solution. 

Let   X   represent  the  greater  number  : 

tnen,  aj— 19   will  represent  the  least, 

and,  2a;— 19  =  67,    whence    2a;  =  67+l9; 

86 
therefore,  x= — =43. 

and  consequently     a;— 19=43  — 19=24. 

General  Solution  of  this  Problem. 

The  sum  of  two  numbers  is   a,  their  difference  is   b. 
What  are  the  two  numbers  ? 


EQUATIONS    OF    THE    FIRST    DEGREE.  81 

Let   X   be  the  least  number, 

x+b   will  represent  the  greater. 
Hence,  2x+bz:^a,   whence   2x=:a—b; 

,        f.  a — b       a       b 

therefore,  X'=.—- — == — , 

'  2  2       2' 

and  consequently,    x-{-bz=i— —+£>=—+—. 

2        <w  2        2 

As  the  form  of  these  two  results  is  independent  of  any 
value  attributed  to  the  letters   a    and    b,  it  follows  that. 

Knowing  the  sum  and  difference  of  two  numbers,  we  obtain 
the  greater  by  adding  the  half  difference  to  the  half  sum,  and 
the  less,  by  subtracting  the  half  difference  from  half  the  sum. 

Thus,  if  the  given  sum  were  237,  and  the  difference  99, 


the  greater  is 

237      99 
2    +2' 

237  +  99       336 
^^2=2 

r=168; 

and  the  least 

237     99 

2         2' 

„   'f  =e. 

Verif  cation. 

168  +  69=237     and     168-69=99. 

8.  A  person  engaged  a  workman  for  48  days.  For 
each  day  that  he  laboured  he  received  24  cents,  and  for 
each  day  that  he  was  idle,  he  paid  12  cents  for  his  board. 
At  the  end  of  the  48  days,  the  account  was  settled,  when 
the  labourer  received  504  cents.  Required  the  number  of 
working  days,  and  the  number  of  days  he  was  idle. 


82  FIRST  LESSONS  IN  ALGEBRA. 

If  these  two  numbers  were  known,  by  multiplying  them 
respectively  by  24  and  12,  then  subtracting  the  last  product 
from  the  first,  the  result  would  be  504.  Let  us  indicate 
these  operations  by  means  of  algebraic  signs. 

Let  X  =    the  number  of  working  days. 

48— a?  =    the  number  of  idle  days. 
Then,        24xx  =z    the  amount  earned, 
and       12(48  —  0?)=:    the  amount  paid  for  his  board. 
Then,       24a:  — 12(48— a:)  =:  504 

what  he  received,  which  is  the  statement.     Then  to  find  x, 
we  first  multiply  by  12,  which  gives 

24a?-576  +  12a:=504. 
or,  36a:=:504+576=:1080, 

x=.  —30   the  workmg  days, 

whence,  48  —  30  =  18   the  idle  days. 

Verification, 

Thirty  day's  labour,  at  24  cents 
a  day,  amounts  to 30x24=720  cents. 

And  18  day's  board,  at  12  cents 

a  day,  amounts  to 18  X  12=216  cents. 

The  difference  is  the  amount  received  504  cents. 

General  Solution. 

This  question  may  be  made  general,  by  denoting  the 
whole  number  of  working  and  idle  days  by   n. 

The  amount  received  for  each  day  he  worked  by  a. 
The  amount  paid  for  his  board,  for  each  idle  day,  by   b. 


EQUATIONS    OF    THE    FIRST    DEGREE.  83 

And  the  balance  due  the  laborer,  or  the  result  of  the 
account,  by   c. 

As  before,  let  the  number  of  working  days  be  repre- 
sented by   X. 

The  number  of  idle  days  will  be  expressed  by   n—x. 

Hence,  what  he  earns  will  be  expressed  by   ax. 

And  the  sum  to  be  deducted,  on  account  of  his  board,  by 

The  equation  of  the  problem  therefore  is 

ax  —  h(n — x^^=zc^ 

which  is  the  statement.     To  find  x  we  first  multiply  by  &, 
which  gives 

ax—hn-{-hx^=zc 
or,  (a4-^)^=c  -{-hn 

whence,  x— -7—=     working  days. 

a  -\-o 

,  ,  c    -\-hn        an-\-hn — c — hn 

and  consequently,  n—x=:n- 


a  +h  a-\-h 

an — c 

a-\-b 


idle  days. 


Let  us  now  suppose  71=48,  «=:24,  5=12,  and  c=504. 
These  numbers  will  give  for  x  the  same  value  as  before 
found. 

9.  A  person  dying  leaves  half  of  his  property  to  his  wife, 
one-sixth  to  each  of  two  daughters,  one-twelfth  to  a  servant, 
and  the  remaining  $600  to  the  poor :  what  was  the  amount 
of  his  property  ^ 


84  FIRST  LESSONS  IN  ALGEBRA. 

Represent  the  amount  of  the  property  by  x. 

X 

Then,         -7^=         what  he  left  to  his  wife, 

— =         Avhat  he  left  to  one  daughter, 

and  — =z —     what  he  left  to  both  daughters, 

6       3 

— =         what  he  left  to  his  servant. 
12 

$600         to  the  poor. 
Then,  by  the  conditions  of  the  question 

X      ,     X      ,     X     ^    ^^^ 

the  amount  of  the  property,  which  gives     a?  =$7200. 

10.  A  and  B  play  together  at  cards.  A  sets  down  with 
$84  and  B  with  $48.  Each  loses  and  wins  in  turn,  when 
it  appears  that  A  has  five  times  as  much  as  B.  How  much 
did  A  win  ? 

Let  X  represent  what  A  won. 
Then  A     rose  with     $84+ a;     dollars, 

and  B     rose  with     $48— a;     dollars. 

But  by  the  conditions  of  the  question,  we  have 
84+a:=5(48-a;), 
hence,  84+a:=240  — 5a: ; 

and,  6a?  =  156, 

consequently,  a? =$26     what  A  won. 

Verijication. 

84+26  =  110;     48-26=22; 

110=5(22)  =  110 


EQUATIONS    OF    THE    FIRST    DEGREE.  85 

11.  A  can  do  a  piece  of  work  alone  in  10  days,  B  in  13 
days  :  in  what  time  can  they  do  it  if  they  work  together  ? 

Denote  the  time  by  x,  and  the  work  to  be  done  by  1. 

Then  in  1  day  A  could  do     — -     of  the  work,  and  B  could 
•^  10 

1  X 

do     —  ;  and  in  x  days  A  could  do     -—     of  the  work,  and 
13'  ^  10 

B)    —  :     hence,  by  the  conditions  of  the  question 

— +— -1- 

10^13-   ' 

which  gives  13a;+10a:=130: 

130 
hence,  23a:=130,     a?—  ~5^     days. 

12.  A  fox,  pursued  by  a  greyhound,  has  a  start  of  60 
leaps.  He  makes  9  leaps  while  the  greyhound  makes  but 
6  ;  but,  three  leaps  of  the  greyhound  are  equivalent  to  7 
of  the  fox.  How  many  leaps  must  the  greyhound  make  to 
overtake  the  fox  ? 

From  the  enunciation,  it  is  evident  that  the  distance  to 
be  passed  over  by  the  greyhound  is  composed  of  the  60 
leaps  which  the  fox  is  in  advance,  plus  the  distance  that  the 
fox  passes  over  from  the  moment  when  the  greyhound  starts 
in  pursuit  of  him.  Hence,  if  we  can  find  the  expression  for 
these  two  distances,  it  will  be  easy  to  form  the  equation  of 
the  problem. 

Let  X  =z  the  number  of  leaps  made  by  the  greyhound 
before  he  overtakes  the  fox. 

Now,  since  the  fox  makes  9  leaps  while  the  greyhound 

9  3 

makes  but  6,  the  fox  will  make     -^     or    — -     leaps  while 

8 


86  FIRST    LESSONS    IN    ALGEBRA. 

the  greyhound  makes  1  ;  and,  therefore,  while  the  greyhound 

3 
makes  x  leaps,  the  fox  will  make     —x    leaps. 

Hence,  the  distance  which  the  greyhound  must  pass  over 

3 

will  be  expressed  by     60-\ x    leaps  of  the  fox. 

It  might  be  supposed,  that  in  order  to  obtain  the  equation, 

3 
it  would  be  sufficient  to  place  x  equal  to     60+— a: ;     but 

in  doing  so,  a  manifest  error  would  be  committed ;  for  the 

leaps  of  the  greyhound  are  greater  than  those  of  the  fox,  and 

we  should  then  equate  numbers  of  different  denominations  ; 

that  is,  numbers  referring  to  different  units.     Hence  it  is 

necessary  to  express  the  leaps  of  the  fox  by  means  of  those 

of  the  greyhound,  or  reciprocally.     Now,  according  to  the 

enunciation,  3  leaps  of  the  greyhound  are  equivalent  to  7 

leaps  of  the  fox,  then  1  leap  of  the  greyhound  is  equiva- 

7 
lent  to     —     leaps  of  the  fox,  and  consequently  x  leaps  of 

7x 
the  greyhound  are  equivalent  to     —     of  the  fox. 

o 

Hence,  we  have  the  equation 

making  the  denominators  disappear 

14a:=:360  +  9.r, 

whence  5x—360     and     a:r=72. 

Therefore  the  greyhound  will  make   72  leaps  to  overtake 
the  fox,  and  during  this  time  the  fox  will  make 

3 
72  X—     or     108. 


y^^\  \  li  iA  A  , ,  ^ 
OF  THE 

UNIVERSITY    3 

,^  Of  / 

EQUATIONS    OF    THE    FIRST    MJCrflEET  87 


Verification. 
The  72  leaps  of  the  greyhound  are  equivalent  to 

leaps  of  the  fox.     And 

60+108  =  168, 

the  leaps  which  the  fox  made  from  the  beginning. 

13.  A  father  leaves  his  property,  amounting  to  $2520,  to 
four  sons,  A,  B,  C,  and  D.  C  is  to  have  $360,  B  as  much 
as  C  and  D  together,  and  A  twice  as  much  as  B  less 
$1000  :  how  much  does  A,  B,  and  D  receive  ? 

Ans,  A  $760,  B  $880,  D  $520. 

14.  An  estate  of  $7500  is  to  be  divided  between  a  widow, 
two  sons,  and  three  daughters,  so  that  each  son  shall  receive 
twice  as  much  as  each  daughter,  and  the  widow  herself 
$500  more  than  all  the  children :  what  was  her  share,  and 
what  the  share  of  each  child  ? 

i  Widow's  share  $4000. 
Ans.  \  Each  son's  $1000. 

V  Each  daughter's  $500. 

15.  A  company  of  180  persons  consists  of  men,  women, 
and  children.  The  men  are  8  more  in  number  than  the 
women,  and  the  children  20  more  than  the  men  and  women 
together  :  how  many  of  each  sort  in  the  company  ? 

Ans.  44  men,  36  women,  100  children. 

16.  A  father  divides  $2000  among  five  sons,  so  that  each 
elder  should  receive  $40  more  than  his  next  younger  bro- 
ther :  what  is  the  share  of  the  youngest  ?  Ans.  $320. 

17.  A  purse  of  $2850  is  to  be  divided  among  three  per- 
sons, A,  B,  and  C.     A's  share  is  to  be  to  B's  as  6  to  11. 


4 


88  FIRST  LESSONS  IN  ALGEBRA. 

and  C  is  to  have  $300  more  than  A  and  B  together :  what 
is  each  one's  share  ?  . 

A71S.  A's  $450,  B's  $825,  C's  $1575. 

18.  Two  pedestrians  start  from  the  same  point ;  the  first 
steps  twice  as  far  as  the  second,  but  the  second  makes  5 
steps  while  the  first  makes  but  one.  At  the  end  of  a  certain 
time  they  are  300  feet  apart.  Now,  allowing  each  of  the 
longer  paces  to  be  3  feet,  how  far  will  each  have  travelled? 

Ans.   1st,  200/eei;  2nd,  500. 

19.  Two  carpenters,  24  journeymen,  and  8  apprentices, 
received  at  the  end  of  a  certain  time  $144.  The  carpen- 
ters received  $1  per  day,  each  journeyman  half  a  dollar, 
and  each  apprentice  25  cents  :  how  many  days  were  they 
employed?  Ans.  9  dai/s, 

20.  A  capitalist  receives  a  yearly  income  of  $2940  :  four- 
fifths  of  his  money  bears  an  interest  of  4  per  cent,  and  the 
remainder  at  5  per  cent :  how  much  has  he  at  interest  ? 

Ans.  70000. 

21.  A  cistern  containing  60  gallons  of  water  has  three 
unequal  cocks  for  discharging  it ;  the  largest  will  empty  it 
in  one  hour,  the  second  in  two  hours,  and  the  third  in  three  : 
in  what  time  will  the  cistern  be  emptied  if  tljey  all  run 
together?  Ans.  32^jmin. 

22.  In  a  certain  orchard  ^  are  apple  trees,  ^  peach  trees, 
^  plum  trees,  120  cherry  trees,  and  80  pear  trees  :  how 
many  trees  in  the  orchard  ?  Ans.  2400. 

23.  A  farmer  being  asked  how  many  sheep  he  had, 
answered  that  he  had  them  in  five  fields;  in  the  1st  he 
had  i,  in  the  2nd  ^,  in  the  3rd  -l-,  and  in  the  4th  ^2^  ^^^  i^ 
the  5th  450  :  how  many  had  he  ?  Ans.  1200. 

24.  My  horse  and  saddle  together  are  worth  $132,  and 
the  horse  is  worth  ten  times  as  much  as  the  saddle  :  what 
is  the  value  of  the  horse  ?  Ans.  120 


EQUATIONS    OF    THE    FIRST    DEGREE.  89 

25.  The  rent  of  an  estate  is  this  year  8  per  cent  greater 
than  it  was  last.  This  year  it  is  $1890  :  what  was  it  last 
year?  Ans.  $1750, 

26.  What  number  is  that  from  which,  if  5  be  subtracted, 
I  of  the  remainder  will  be  40  1  Ans.  65, 

27.  A  post  is  i  in  the  mud,  J  in  the  water,  and  10  feet 
above  the  water  :  what  is  the  whole  length  of  the  post  ? 

Ans.  24  feet. 

28.  After  paying  ^  and  ^  of  my  money,  I  had  66  guineas 
left  in  my  purse  :  how  many  guineas  were  in  it  at  first  ? 

Ans.  120. 

29.  A  person  was  desirous  of  giving  3  pence  apiece  to 
some  beggars,  but  found  he  had  not  money  enough  in  his 
pocket  by  8  pence :  he  therefore  gave  them  each  2  pence 
and  had  3  pence  remaining :  required  the  number  of  beggars. 

Ans.  11. 

30.  A  person  in  play  lost  i  of  his  money,  and  then  won 
3  shillings  ;  after  which  he  lost  J  of  what  he  then  had ; 
and  this  done,  found  that  he  had  but  12  shillings  remaining  : 
what  had  he  at  first  ?  Ans.  20s. 

31.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money 
in  trade  ;  A  gains  $126,  and  B  loses  $87,  and  A's  money 
is  now  double  of  B's  :  what  did  each  lay  out  ? 

Ans.  $300. 

32.  A  person  goes  to  a  tavern  with  a  certain  sum  of  mo- 
ney in  his  pocket,  where  he  spends  2  shillings  ;  he  then 
borrows  as  much  money  as  he  had  left,  and  going  to  another 
tavern,  he  there  spends  2  shillings  also ;  then  borrowing 
again  as  much  money  as  was  left,  he  went  to  a  third  tavern, 
where  likewise  he  spent  two  shillings  and  borrowed  as 
much  as  he  had  left ;  and  again  spending  2  shillings  at  a 
fourth  tavern,  he  then  had  nothing  remaining.  What  had 
he  at  first  ?  Ans.  3s.  9d. 

8* 


90  FIRST    LESSONS    IN    ALGEBRA. 


Of  Equations  of  the  First  Degree  involving  two  or 
more  unknown  quantities, 

72.  Although  several  of  the  questions  hitherto  resolved 
contained  in  their  enunciation  more  than  one  unknown  quan- 
tity, we  have  resolved  them  all  by  employing  but  one  sym- 
bol. The  reason  of  this  is,  that  we  have  been  able,  from 
the  conditions  of  the  enunciation,  to  express  easily  the  other 
unknown  quantities  by  means  of  this  symbol ;  but  we  are 
unable  to  do  this  in  all  problems  containing  more  than  one 
unknown  quantity. 

To  ascertain  how  problems  of  this  kind  are  resolved,  let 
us  take  some  of  those  which  have  been  resolved  by  means 
of  one  unknown  quantity. 

1.  Given  the  sum  of  two  numbers  equal  to  36  and  their 
difference  equal  to  12,  to  find  the  numbers. 

Let  X—  the  greater,  and  y=z  the  less  number. 

Then,  by  the  1st  condition a:+y=3G, 

and  by  the  2nd,        x—yz=:l2. 

By  adding  (Art.  65,  Ax.  1),       ....  2a:i=48. 

By  subtracting  (Art.  65,  Ax.  2),     .     .     .  2y—24:. 

Each  of  these  equations  contains  but  one  unknown  quantity. 

48 
From  the  first  we  obtain x———2i. 

24 

And  from  the  second ?/=— -=j:12. 

2 

Verification, 

a:+y  =  36     gives     24+12  =  36. 
^-y=12        „        24  —  12  =  12. 


EQUATIONS    OF    THE    FIRST    DEGREE.  91 

General  Solution. 
Let  x^=z  the  greater,  and  y  the  less  number. 

Then  by  the  conditions a;+y=a, 

and         x—yzzzh. 

By  adding,  (Art.  65,  Ax.  1),      .     .     .  2x=a+h. 

By  subtracting,  (Art.  65,  Ax.  2),    .     .  2y  =  a-^h. 

Each  of  these  equations  contains  but  one  unknown  quantity* 

a+h 
From  the  first  we  obtain x=:—~ — . 

a — b 
And  from  the  second y  =:——-. 

^        2 

Verification, 

a-{-b      a  —  h      2a  .     a-\-h      a—h      2b 

= —  =  a:     and     — = — =zb. 

2^22'  2  22 

For  a  second  example,  let  us  also  take  a  problem  that  has 
been  already  solved. 

2.  A  person  engaged  a  workman  for  48  days.  For  each 
day  that  he  labored  he  was  to  receive  24  cents,  and  for  each 
day  that  he  was  idle  he  was  to  pay  12  cents  for  his  board. 
At  the  end  of  the  48  days  the  account  was  settled,  when 
the  laborer  received  504  cents.  Required  the  number  of 
working  days,  and  the  number  of  days  he  was  idle. 

Let  x=         the  number  of  working  days, 

y=         the  number  of  idle  days. 
Then,         24x=         what  he  earned, 
and  12y=         what  he  paid  for  his  board. 

Then,  by  the  conditions  of  the  question,  we  have 

x-^y     =48, 
and  24a;— 12y=:504. 

This  is  the  statement  of  the  question. 


92  FIRST    LESSONS    IN    ALGEBRA. 

It  has  already  been  shown  (Art.  65,  Ax.  3),  that  the  two 
members  of  an  equation  can  be  multiplied  by  the  same  num- 
ber, without  destroying  the  equality.  Let,  then,  the  first 
equation  be  multiplied  by  24,  the  coefficient  of  x  in  the 
second  :  we  shall  then  have 

24a;+24yz=1152, 

24a;— 12y—   504. 

And  by  subtracting,  36y=z   648, 

and  y  =  -3^=-l^- 

Substituting  this  value  of  y  in  the  equation 

24a;— 12y  =  504,     we  have     24a;— 216  =  504, 

which  gives 

720 
24a:==:504  +  216zi:720,     and     xz=: — —=30. 

24 

Verification. 

a;+     y—  48     gives  30+18=  48, 

24a:— 12y=504     gives     24x30  —  12x18  =  504. 

Elimination. 

73.  The  method  which  has  just  been  explained  of  com- 
bining two  equations,  involving  two  unknown  quantities,  and 
deducing  therefrom  a  single  equation  involving  but  one,  is 
called  elimination. 

Quest. — 73.  What  is  elimination  1  How  many  methods  of  elimina- 
tion are  there  1  Give  the  rule  for  elimination  by  addition  and  subtrac- 
tion ^     What  is  the  first  stepi     What  the  second  ?     What  the  third  ] 


EQUATIONS    OF    THE    FIRST    DEGREE.  93 

There  are  three  principal  methods  of  elimination : 

1st.  By  addition  and  subtraction. 

2d.    By  substitution. 

3d.    By  comparison. 
We  will  consider  these  methods  separately. 

Elimination  hy  Addition  and  Subtraction. 

1.  Take  the  two  equations 

Sic— 2y=7 

8a;+2y=48. 

If  we  add  these  two  equations,  member  to  member,  we 
obtain 

lla;=55. 
which  gives,  by  dividing  by  11 

x=:5  : 

and  substituting  this  value  in  either  of  the  given  equations, 
we  find 

2.  Again,  take  the  equations 

8a;+2y=z48 
3x+2j/z=23. 

If  we  subtract  the  2nd  equation  from  the  first,  we  obtain 
5xz=z25, 
which  gives,  by  dividing  by  5 

x=z5  : 
and  by  substituting  this  value,  we  find 

y=4. 


94  FIRST    LESSONS    IN   ALGEBRA. 

3.  Take  the  two  equations 

lla;+9y=69. 

If,  in  these  equations,  one  of  the  unknown  quantities  was 
affected  with  the  same  coefficient,  we  might,  by  a  simple 
subtraction,  form  a  new  equation  which  would  contain  but 
one  unknown  quantity. 

Now,  if  both  members  of  the  first  equation  be  multiplied 
by  9,  the  coefficient  of  y  in  the  second,  and  the  two  mem- 
bers of  the  second  by  7,  the  coefficient  of  y  in  the  first,  we 
will  obtain 

45x+63y  =  387, 
77a:+63yr=483. 

Subtracting,  then,  the  first  of  these  equations  from  the 
second,  there  results 

32a;=:96,     whence     x=z3. 

Again,  if  we  multiply  both  members  of  the  first  equation 
by  11 ,  the  coefficient  of  x  in  the  second,  and  both  members 
of  the  second  by  5,  the  coefficient  of  x  in  the  first,  we  will 
form  the  two  equations 

55xi-77i/  =  473, 
55a?+45y=:345. 

Subtracting,  then,  the  second  of  these  two  equations  from 
the  first,  there  results 

32y=128,     whence     y  =  4. 
Therefore  x=z3  and  y=4,  are  the  values  of  x  and  y. 

Verification. 

5x+7y=43     gives       5x3  +  7x4=zl5+28==43 ; 

lla;+95=:69        „         11  x3  +  9x4=:334-36=69. 


EQUATIONS    OF    THE    FIRST    DEGREE.  95 

The  method  of  elimination  just  explained,  is  called  the 
method  by  addition  and  subtraction. 

To  eliminate  by  this  method  we  have  the  following 

RULE. 

I.  See  which  of  the  unknown  quantities  you  will  eliminate. 

II.  Make  the  coefficient  of  this  unknown  quantity  the  same 
in  both  equations,  either  by  multiplication  or  division. 

III.  If  the  signs  of  the  like  terms  are  the  same  in  both 
equations,  subtract  one  equation  from  the  other ;  but  if  the 
signs  are  unlike,  add  them. 

j:XAMPLES. 

4.  Find  the  values  of  x  and  y  in  the  equations 

3a; — y  =  3, 
y-\-2x=:7. 

Ans.  xz=2,  y=z3 

5.  Find  the  values  of  x  and  y  in  the  equations 

4x-7y:=^  -22, 
5x+2y=37. 

Ans.  x=z5,  y=:6, 

6.  Find  the  values  of  x  and  y  in  the  equations 

2x+6y  =  42, 
8x—6y=  3. 

Ans.  37=4^,  y=z5^. 

7.  Find  the  values  of  x  and  y  in  the  equations 

8x~9y=l. 
6x^3y=4x. 

Ans.  x=^,  y=J, 


96  FIRST    LESSONS    IN    ALGEBRA. 

8,  Find  the  values  of  x  and  y  in  the  equations 

14ir— 15y=:12, 
lx+   8y=37. 

Ans.  a:r=3,  y=2. 

9,  Find  the  values  of  x  and  y  in  the  equations 

^77^.  37=6,  y=:9. 

10,  Find  the  values  of  x  and  y  in  the  equations 


\ 


— ^+— y=6i- 


a?— y=z:— 2. 

Ans.  x=z\4:,  y=16. 

11.  Says  A  to  B,  you  give  me  $40  of  your  money,  and 
I  shall  then  have  5  times  as  much  as  you  v^ill  have  left. 
Now  they  both  had  $120  :  how  much  had  each  ? 

Ans.  Each  had  $60. 

12.  A  Father  says  to  his  son,  "  twenty  years  ago,  my 
age  was  four  times  yours  ;  now,  it  is  just  double  :"  what  were 
their  ages  1  .  (  Father's  60  years. 

)  Son's       30  years. 

13.  A  Father  divides  his  property  between  his  two  sons. 
At  the  end  of  the  first  year  the  elder  had  spent  one  quarter 
of  his,  and  the  younger  had  made  $1000,  and  their  property 
was  then  equal.  After  this  the  elder  spends  $500  and  the 
younger  makes  $2000,  when  it  appears  the  younger  has 
just  double  the  elder  :  what  had  each  from  the  father  ? 

.         5  Elder        $4000. 
i  Younger  $2000. 


EQUATIONS    OF    THE    FIRST    DEGREE.  97 

14.  If  John  give  Charles  15  apples,  they  will  have  the 
same  number;  but  if  Charles  give  15  to  John,  John  will 
have  15  times  as  many  wanting  10  as  Charles  will  have 
left.     How  many  had  each  J  .         (  John       50. 

(  Charles  20. 

15.  Two  clerks,  A  and  B,  have  salaries  which  are  to- 
gether equal  to  $900.  A  spends  -jL-  per  year  of  what  he 
receives,  and  B  adds  as  much  to  his  as  A  spends.  At  the 
end  of  the  year  they  have  equal  sums  :  what  was  the  salary 

of  each?  .         (  A's=:500. 

Ans.    < 

(  B's=400. 


p^  Elimination  by  Substitution. 

7  4.  Let  us  again  take  the  equations 
5a;+7y=:43, 

Find  the  value  of  x  in  the  first  equation,  which  gives 


Substitute  this  value  of  x  in  the  second  equation,  and  we 
have 


J.X/N              -r^!/  —  ^^f 

or, 

473-77y+45y  =  345, 

or, 

-32^  =  — 128. 

Hence, 

y=4, 

and. 

43-28     ^ 

^       ■    5          "*• 
9 

98  FIRST    LESSONS    IN    ALGEBRA. 

This  method  is  called  the  method  by  substitution:  we 
have  for  it  the  following 

RULE. 

Find  the  value  of  one  of  the  unknovyn  quantities  in  either 
of  the  equations,  and  substitute  this  value  for  the  same  unknown 
quantity  in  the  other  equation :  there  will  thus  arise  a  new 
equation  with  but  one  unknown  quantity. 

Remark. — This  method  of  elimination  is  used  to  great 
advantage  when  the  coefficient  of  either  of  the  unknown 
quantities  is  unity. 

EXAMPLES. 

1.  Find,  by  the  last  method,  the  values  of  x  and  y  in  the 
equations 

3a?— y=:l      and     3y— 2ac  =  4 

Ans.  x:=z\,  y=.2. 

2.  Find  the  values  of  x  and  y  in  the  equations 

by  — 4.XZZ1— 22     and     3y  +  4a:  =  38. 

Ans.  a?i=:8,  y=2. 

3.  Find  the  values  of  x  and  y  in  the  equations 

x^Sy  —  \Q     and     y~2xz=:z—29. 

Ans.  a?=zlO,  y=l. 

4.  Find  the  values  of  x  and  y  in  the  equations 

2 
5a;— y=:13    and    8a;+— j/=:29. 

Ans.  a: =3 J,  yz=zA^. 

Quest. — 74.  Give  the  rule  for  elimination  by  substitution  '^  "Wlien  is 
it  desirable  to  use  this  method  1 


EQUATIONS    OF    THE    FIRST    DEGREE.  09 

5.  Find  the  values  of  x   and   y   from  the  equations 

10a?— —=69     and     lOy— — =49. 

Ans.  x=:7,  y=:5. 

6.  Find  the  values  of  x   and   y   from  the  equations 

^y^^.  07=8,  y=10. 

7.  Find  the  values  of  x  and   y   in  the  equations 


-^-1  +  5  =  3,    .+X^17i. 

Ans.  x=zl5,  y=14. 
8    Find  the  values  of  x  and   y   in  the  equations 

^/ii".  a?  =  3J,  y  =  4. 
9.  Find  the  values  of  x  and   y   from  the  equations 

4 h6  =  5     and ~=0. 

8       4^  12      16 

V—  Ans,  ir=12,  y=16. 

10.  Find  the  values  of  x   and   y    from  the  equations 

^-|-l  =  -9     and     5.-1=29. 

Ans.  x=:6j  y=:7. 

11.  Two  misers  A  and  B  sit  down  to  count  over  their 
money.  They  both  have  $20000,  and  B  has  three  times  as 
much  as  A  :  how  much  has  each  ? 

A  .  .       $5000. 
$15000. 


Ans.  \  g  ; 


FIRST  LESSONS  IN  ALGEBRA. 

A  person  has  two  purses.  If  he  puts  $7  into  the 
first  purse,  it  is  worth  three  times  as  much  as  ihe  second : 
but  if  he  puts  $7  into  the  second  it  becomes  worth  five 
times  as  much  as  the  first :  what  is  the  value  of  each  purse  ? 

Ans.   1st,  $2:  2nd,  $3. 

13.  Two  numbers  have  the  following  properties  :  if  the 
first  be  multiplied  by  6  the  product  will  be  equal  to  the 
second  multiplied  by  5  ;  and  one  subtracted  from  the  first 
leaves  the  same  remainder  as  2  subtracted  from  the  second  : 
what  are  the  numbers  ?  Aiis.  5  and  6. 

14.  Find  two  numbers  with  the  following  properties  : 
the  first  increased  by  2  to  be  3^  times  greater  than  the 
second :  and  the  second  increased  by  4  to  be  half  the  first : 
what  are  the  numbers  ?  Ans.  24  and  8. 

15.  A  father  says  to  his  son,  "  twelve  years  ago  I  was 
twice  as  old  as  you  are  now :  four  times  your  age,  at  that 
time,  plus  twelve  years,  will  express  my  age  twelve  years 
hence :"  what  were  their  ages  1        ^       ^  Father  72  years. 


'1 


Son       30 


Elimination  by  Comparison. 

75.  Take  the  same  equations 

5a;+7yr:r43, 

lla:-f9yrz:69. 

Finding  the  value  of  x  in  the  first  equation,  we  have 

43 -7y 
5 

and  finding  the  value  of  x   in  the  second,  we  obtain 

_  69— 9y 
""-      11    / 


EQUATIONS    OF    THE    FIRST    DEGREE.  101 

Let  these  two  values  of  x  be  placed  equal  to  each  other, 
and  we  have 

43— 7y  _  69  — 9y 

5     ~    n    * 

Or,  473-77y==:34-5-45y ; 

Or,  -32y=:-128. 

Hence,  y=4. 

And,  xz=: — — =^3. 

This  method   of  elimination   is    called  the    method  by 
c  )mparison,  for  which  we  have  the  following 

RULE. 

I.  Find  the  value  of  the  same  uiiknown  quantity  in  each 
€^  uation, 

II.  Place  these  values  equal   to  each  other ;  and   a    new 
equation  will  arise  with  hut  one  unknown  quantity.   " 

EXAMPLES. 

1.  Find,  by  the  last  rule,  the  values  of  x   and  y   in  the 
equations 

3x+|-+6=42     and     y— =14^. 

Ans.  a?=ll,  yi=15. 

Quest. — 75.  Give  the  rule  for  elimination  by  comparison  1     What  is 
the  first  step  \     What  the  second  \ 

9* 


102  FIRST    LESSONS    IN    ALGEBRA. 

2.  Find  the  values  of  x   and   y  in  the  equations 

f-f +5^6     and     f +4=^+6. 

/  Ans.  X—2S,  y=20. 

3.  Find  the  values  of  x  and  y  in  the  equations 

-^ =1     and     3y— a:  =  6. 

10      4       8  ^ 

Ans.   a7r=:9,  y=:5. 

4.  Find  the  values  of  x   and   y   in  the  equations 

1  T— I—  y 

y—^  =  ~x+^     and     ^^^-^rry— 3|. 

Ans.  a;=2,  y=9. 

5.  Find  the  values  of  x   and   y   in  the  equations 

V— a?  ,    X  ^  .      X    ,    y 

y_+_=y_2     and     _+X=^_i3. 

^n^.  07=16,  y=:7. 

6.  Find  the  values  of  x   and  y   from  the  equations 

y-\-x  ,  y— ^  2y 

:^ \-^ zzzx — :f,     and     x+y  =  l6. 

2^2  3'  ^ 

Ans.  a:=rlO,  y=:6, 

7.  Find  the  values  of  x  and  y  in  the  equations 

^/t^.  a:=:l,     y=3. 

8.  Find  the  values  of  x  and  y  in  the  equations 

X — 4  a; 

2y+3a^=y+43,    y ^=y-j-. 

Ans.  x:=^lOj    y=13. 


EQUATIONS    OF    THE    FIRST    DEGREE.  103 

9.  Find  the  values  of  x  and  y  in  the  equations 

4i/—^-^^x+l8,    and   27— y  =  a;+y+4. 

Ans.  x=:9,     y=7. 

10.  Find  the  values  of  x  and  y  in  the  equations 

Ans.  a;=10,     y=20. 

76.  Having  explained  the  principal  methods  of  elimina- 
tion, we  shall  add  a  few  examples  which  may  be  solved  by 
either ;  and  often  indeed,  it  may  be  advantageous  to  use 
them  all  even  in  the  same  question. 


GENERAL    EXAMPLES. 

1.  Given     2x+3t/=zl6,    and    3a;— 2y=:ll    to  find  the 
values  of  x  and  y.  Ans.  a?=5,   y=2. 

„    ^.  2x     3y       9         ,    3a?     2y       61         ^    ,   , 

2.  Given    -j+f =3^    and  -j+i^^^  to  find  the 

values  of  x  and  y.  Ans.  x= — ,   y=: — . 

^  o 

3.  Given     -|-+7y=99,   and  -^+7a?=51,    to  find  the 
values  of  x  and  y.  Ans.  a; =7,   y:=:14. 

4.  Given 

to  find  the  values  of  x  and  y.  .Aw^.  aj=60,  y;5:40. 


104  FIRST  lessoj?^s  in  algebra. 


QUESTIONS. 

1.  What  fraction  is  that,  to  the  numerator  of  which,  if  I 
be  added,  its  value  will  be     — ,     but  if  one  be  added  to  its 

denominator,  its  value  will  be     —  ? 


Let  the  fraction  be  represented  by 
Then,  by  the  question 


x+l       1  ^         a;  1 

=:-—     and      — —-=-—. 

y         3  y+1       4 

Whence  3x+3=y     and     4x=zy-\-l. 

Therefore,  by  subtracting, 

ac— 3  =  1     or     x=:4. 
Hence,  12  +  3=y; 

therefore,  y=15. 

2.  A  market-woman  bought  a  certain  number  of  eggs  at 
2  for  a  penny,  and  as  many  others,  at  3  for  a  penny ;  and 
having  sold  them  again  altogether,  at  the  rate  of  5  for  2d, 
found  that  she  had  lost  4d :  how  many  eggs  had  she  ? 

2x=z     the  whole  number  of  eggs. 
a?=     the  number  of  eggs  of  each  sort. 

—oc=     the  cost  of  the  first  sort, 
—x=     the  cost  of  the  second  sort. 

o 

:  2x  :  -—    the  amount  for  whict  the  egg 
5 


Let 

Then 

Then  will 

and 

But     5:2 

were  sold. 

EQUATIONS    OF    THE    FIRST    DEGREE,  105 

Hence,  by  the  question 

Therefore,  15«c+10a:— 24a;=120, 

or  xz=zl20; 

the  number  of  eggs  of  «ach  sort, 

3,  A  person  possessed  a  capital  of  30,000  dollars,  for 
which  he  drew  a  certain  interest ;  but  he  owed  the  sum  of 
20,000  dollars,  for  which  he  paid  a  certain  interest.  The 
interest  that  he  received  exceeded  that  which  he  paid  by 
800  dollars.  Another  person  possessed  35,000  dollars,  for 
which  he  received  interest  at  the  second  of  the  above  rates  ; 
but  he  owed  24,000  dollars,  for  which  he  paid  interest  at 
the  first  of  the  above  rates.  The  interest  that  he  received 
exceeded  that  which  he  paid  by  310  dollars.  Required  the 
two  rates  of  interest. 

Let  X  and  y  denote  the  two  rates  of  interest ;  that  is,  the 
interest  of  $100  for  the  given  time. 

To  obtain  the  interest  of  $30,000  at  the  first  rate,  denoted 
by  X,  we  form  the  proportion 

100  :  X  ::  30,000  :  :    ^^~^      or     300^, 

And  for  the  interest  $20,000,  the  rate  being  y, 

100  :  y  :  :  20,000  :  :    ^^^      or     200y. 

But  from  the  enunciation,  the  difference  between  these 
two  interests  is  equal  to  800  dollars. 

We  have,  then,  for  the  first  equation  of  the  problem, 

300;r— 200y=800. 


106  FIRST  LESSONS  IN  ALGEBRA. 

By  writing  algebraically  the  second  condition  of  the  pro- 
blem, we  obtain  the  other  equation, 

350y  — 240a?=310. 

Both  members  of  the  first  equation  being  divisible  by  1 00, 
and  those  of  the  second  by  10,  we  may  put  the  following, 
in  place  of  them  : 

3a:--2y  =  8,  35y— 24a;=:31. 

To  eliminate  x,  multiply  the  first  equation  by  8,  and  then 
add  it  to  the  second ;  there  results 

19y=:95,     whence     y  =  5. 

Substituting  for  y,  in  the  first  equation,  its  value,  this 
equation  becomes 

3a?— 10  =  8,     whence     a?=:6. 
Therefore,  the  first  rate  is  6  per  cent,  and  the  second  5, 

Yerif  cation. 
$30,000,     placed  at  6  per  cent,  gives     300  x  6  =  $1800. 
$20,000,  „  5       „  „        200x5  =  $1000, 

And  we  have  1 800  —  1 000  =  800. 

The  second  condition  can  be  verified  in  the  same  manner. 

-  4.  "What  two  numbers  are  those,  whose  difference  is  7, 
and  sum  33  ?  Ans.  13  and  20. 

5.  To  divide  the  number  75  into  two  such  parts,  that 
three  times  the  greater  may  exceed  seven  times  the  less 
by  15.  Ans.  54  and  21. 

6.  In  a  mixture  of  wine  and  cider,  i  of  the  whole  plus 
25  gallons  was  wine,  and  ^  part  minus  5  gallons  was  cider ; 
how  many  gallons  were  there  of  each  ? 

Ans.  85  of  wine,  and  35  of  cider. 


EQUATIONS    OF    THE    FIRST    DEGREE.  107 

7.  A  bill  of  jC120  was  paid  in  guineas  and  moidores,  and 
the  number  of  pieces  of  both  sorts  that  were  used  was  just 
100.  If  the  guinea  be  estimated  at  21^,  and  the  moidore 
at  27^,  how  many  were  there  of  each  ?      Ans.  50  of  each. 

8.  Two  travellers  set  out  at  the  same  time  from  London 
and  York,  whose  distance  apart  is  150  miles.  One  of  them 
goes  8  miles  a  day,  and  the  other  7  :  in  what  time  will  they 
meet?  Ans,  In  10  days. 

9.  At  a  certain  election,  375  persons  voted  for  two  can- 
didates, and  the  candidate  chosen  had  a  majority  of  91  : 
how  many  voted  for  each  ? 

Ans,  233  for  one,  and  142  for  the  other. 

10.  A  person  has  two  horses,  and  a  saddle  worth  jC50. 
Now,  if  the  saddle  be  put  on  the  back  of  the  first  horse,  it 
will  make  his  value  double  that  of  the  second ;  but  if  it  be 
put  on  the  back  of  the  second,  it  will  make  his  value  triple 
that  of  the  first.     What  is  the  value  of  each  horse  ? 

Ans.  One  jC30,  and  the  other  £^0, 

1 1 .  The  hour  and  minute  hands  of  a  clock  are  exactly 
together  at  12  o'clock  :  when  are  they  next  together  ? 

Ans.   \hr.  5^min. 

12.  A  man  and  his  wife  usually  drank  out  a  cask  of  beer 
in  12  days ;  but  w^hen  the  man  was  from  home,  it  lasted 
the  woman  30  days  :  how  many  days  would  the  man  alone 
be  in  drinking  it  1  Ans.  20  days. 

13.  If  32  pounds  of  sea-water  contain  1  pound  of  salt, 
how  much  fresh  water  must  be  added  to  these  32  pounds, 
in  order  that  the  quantity  of  salt  contained  in  32  pounds  of 
the  new  mixture  shall  be  reduced  to  2  ounces,  or  |-  of  a 
pound?  Ans.  224lb. 

14.  A  person  who  possessed  100,000  dollars,  placed  the 
greater  part  of  it  out  at  5  per  cent  interest,  and  the  other 


108  FIRST  LESSONS  IN  ALGEBRA. 

at  4  per  cent.     The  interest  which  he  received  for  the  wholo 
amounted  to  4640  dollars.     Required  the  two  parts. 

Ans.  64,000  and  36,000 

15.  At  the  close  of  an  election,  the  successful  candidate 
had  a  majority  of  1500  votes.  Had  a  fourth  of  the  votes 
of  the  unsuccessful  candidate  been  also  given  to  him,  he 
would  have  received  three  times  as  many  as  his  competitor 
wanting  three  thousand  fivehundred  :  how  many  votes  did 

i  2d,    5000. 

16.  A  gentlemen  bought  a  gold  and  a  silver  watch,  and 
a  chain  worth  $25.  When  he  put  the  chain  on  the  gold 
watch,  it  was  worth  three  and  a  half  times  more  than  the 
silver  watch ;  but  when  he  put  the  chain  on  the  silver  watch, 
it  was  worth  one  half  the  gold  watch  and  1 5  dollars  over : 
what  was  the  value  of  each  watch  ? 

Gold  watch  $80. 


each  receive  ?  ,         Ost,  6500. 

Ans. 


Ans.     ,  _., 

Silver     „      $30. 

17.  There  is  a  certain  number  expressed  by  two  figureSy 
which  figures  are  called  digits.  The  sum  of  the  digits  is 
11,  and  if  13  be  added  to  the  first  digit  the  sum  will  be  three 
times  the  second  :  what  is  the  number  ?  Ans.  56. 

18.  From  a  company  of  ladies  and  gentlemen  15  ladies 
retire ;  there  are  then  left  two  gentlemen  to  each  lady. 
After  which,  45  gentlemen  depart,  when  there  are  left  5 
ladies  to  each  gentleman  :  how  many  were  there  of  each  at 
first?  .         5  50  gentlemen. 

i  40  ladies. 

19.  A  person  wishes  to  dispose  of  his  horse  by  lottery. 
If  he  sells  the  tickets  at  $2  each,  he  will  lose  $30  on  his 
horse  ;  but  if  he  sells  them  at  $3  each,  he  will  receive  $30 
more  than  his  horse  cost  him.  What  is  the  value  of  the 
horse  and  number  of  tickets  ?       .         (Horse    .  .  .  $150. 

No.  of  tickets     60. 


'■\ 


EQUATIONS    OF    THE    FIRST    DEGREE.  109 

20.  A  person  purchases  a  lot  of  wheat  at  $1 ,  and  a  lot  of 
rye  at  75  cents  per  bushel,  the  whole  costing  him  $117,50. 
He  then  sells  ^  of  his  wheat  and  ^  of  his  rye  at  the  same 
rate,  and  realizes  $27,50.     How  much  did  he  buy  of  each  ? 

]  80bu.  of  wheat. 


Ans.     .       ,        ^ 

50bu.  of  rye. 


Equations  involving  three  or  more  unknown  quantities, 

77.  Let  us  now  consider  the  case  of  three  equations 
involving  three  unknown  quantities. 
Take  the  equations 

5.T— 6y+4;sr=15, 
7a:+4y  — 3^=19, 
2.t4-  y+6;s:=:46. 

To  eliminate  z  by  means  of  the  first  two  equations,  mul- 
tiply the  first  by  3  and  the  second  by  4 ;  then,  since  the 
coefiicients  of  z  have  contrary  signs,  add  the  two  results 
together.     This  gives  a  new  equation  : 

43a:-2y=:121. 

Multiplying  the  second  equation  by  2,  a  factor  of  the  co- 
efficient of  z  in  the  third  equation,  and  adding  them  together, 
we  have 

16a;-f9y=:84. 

The  question  is  then  reduced  to  finding  the  values  of  x 
and  y,  which  will  satisfy  these  new  equations. 

Now,  if  the  first  be  multiplied  by  9,  the  second  by  2,  and 
the  results  be  added  together,  we  find 

4 19a: =1257,     whence     x^Z. 
10 


110  FIRST    LESSONS    IN    ALGEBIIA. 

"We  might,  by  means  of  the  two  equations  involving  x 
and  y,  determine  y  in  the  same  way  we  have  determined  x ; 
but  the  value  of  y  may  be  determined  more  simply,  by  ob- 
serving that  the  last  of  these  two  equations  becomes,  by 
substituting  for  x  its  value  found  above, 

84—48 
48-}-9y=:84,     whence    y  =^ =:4. 

In  the  same  manner  the  first  of  the  three  proposed  equa- 
tions becomes,  by  substituting  the  values  of  x  and  y, 

24 
15 — 24-\-4:Zz=:l5,     whence     2;=z —  =  6. 

Hence,  to  solve  equations  containing  three  or  more  un- 
known quantities,  we  have  the  following 


RULE. 

I.  To  eliminate  one  of  the  unknown  quantities,  combine  any 
one  of  the  equations  with  each  of  the  others ;  there  will  thus  he 
obtained  a  series  of  new  equations  containing  one  less  unknown 
quantity, 

II.  Eliminate  another  unknown  quantity  by  combining  one 
of  these  new  equations  with  the  others. 

III.  Continue  this  series  of  operations  until  a  single  equa- 
tion containing  but  one  unknown  quantity  is  obtained,  from 
which  the  value  of  this  unknown  quantity  is  easily  found. 
Then,  by  going  back  through  the  series  of  equations  which  have 
been  obtained,  the  values  of  the  other  unknown  quantities  may 
he  successively  determined. 

Quest. — 77.  Give  the  general  rule  for  solving  equations  involving 
three  or  more  unknown  quantities  1  What  is  the  first  step  T  What  the 
second  1     What  the  third  1 


EQUATIONS    OF    THE    FIRST    DEGREE.  Ill 

78.  Remark. — It  often  happens  that  each  of  the  pro- 
posed equations  does  not  contain  all  the  unknown  quantities. 
In  this  case,  with  a  little  address,  the  elimination  is  very 
quickly  performed. 

Take  the  four  equations  involving  four  unknown  quantities : 

(1.)     2x  —  3i/-i-2z=:'[3.  (3.)     4i/+2zz=zl4. 

(2.)  4u—2xz=30.  (4.)     5y+3w=:32. 

By  inspecting  these  equations,  we  see  that  the  elimina- 
tion of  z  in  the  two  equations,  (1)  and  (3),  will  give  aa 
equation  involving  x  and  y ;  and  if  we  eliminate  u  in 
the  equations  (2)  and  (4),  we  shall  obtain  a  second  equation, 
involving  x  and  y.  These  two  last  unknown  quantities 
may  therefore  be  easily  determined.  In  the  first  place,  the 
elimination  of  z   in  (1)  and  (3)  gives 

7y— 2a:~l  ; 

That  of  u   in  (2)  and  (4),  gives 

207/-\-6x  =  3S. 

Multiplying  the  first  of  these  equations  by  3,  and  adding, 

41y-41  ; 

Whence  y—    1. 

Substituting  this  value  in    7y— 2a:=l,  we  find 

a:=3. 

Substituting  for  x  its  value  in  equation  (2),  it  becomes 

Au—6=z30: 

Whence  u  =  9. 

And  substituting  for   y   its  value  in  equation  (3),  there 

results  z=z5. 


112 


FIRST  LESSONS  IN  ALGEBRA. 


1.  Given    -^ 


EXAMPLES. 
X+       y+       2-=:29' 

x+   2y+    3^=62 


>  to  find  a',  y  and  2^. 


^«5.      a?=:r8,  y=:9,   ;^— 12. 

2a; -f   4y—   3^=22"] 

2.  Given    ^     4a;—  2y4-   5^^=18  ^  to  find  a?,  y  and  z, 

I   6x+  7ij—     z=:63j 

Arts.     a?=i:3,  y=7,  ^  =  4. 

^+-2-y+Y^=:32 

3.  Given    ^  — a:-h— -3/+— 2^— 15  ^  to  find  a?,  y  and  z. 

AnS.       Xz^\2,    y=:20,    ;?r=:30. 

4.  Divide  the  number  90  into  four  sucli  parts  that  the 
first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  be  equal 
to  each  other. 

This  question  may  be  easily  solved  by  introducing  a  new 
unknown  quantity. 

Let  X,  y,  z,  and  u,  be  the  required  parts,  and  desig- 
nate by  m  the  several  equal  quantities  which  arise  from 
the  conditions.     We  shall  then  have 


'<4 


EQUATIONS    OF    THE    FIRST    DEGREE.  113 

From  which  we  find 

m 
x=:m — 2,    yr=m+2,   z=:—,  u=z2m. 

And  by  addmg  the  equations, 

771 

x-\-i/-\-z-\-u=^m-\-m-\ — —-\-2m=z4^m. 

And  since,  by  the  conditions  of  the  question,  the  first 
member  is  equal  to  90,  we  have 

4j77?=z90,     or     |w=90; 
hence  mr=20. 

Having  the  value  of  m^  we  easily  find  the  other  values  : 
viz. 

a;=18,    y:=22,    z—\^,    w=40.  ^^ 

5.  There  are  three  ingots  composed  of  different  metals 
mixed  together.  A  pound  of  the  first  contains  7  ounces  of 
silver,  3  ounces  of  copper,  and  6  of  pewter.  A  pound  of 
the  second  contains  12  ounces  of  silver,  3  ounces  of  cop- 
per, and  1  of  pewter,  A  pound  of  the  third  contains  4 
ounces  of  silver,  7  ounces  of  copper,  and  5  of  pewter.  It 
is  required  to  find  how  much  it  will  take  of  each  of  the 
three  ingots  to  form  a  fourth,  which  shall  contain  in  a 
pound,  8  ounces  of  silver,  3f  of  copper,  and  4^  of  pewter. 

Let  ir,  y,  and  z  represent  the  number  of  ounces  which  it 

is  necessary  to  take  from  the  three  ingots  respectively,  in 

order  to  form  a  pound  of  the  required  ingot.     Since  there 

are  7  ounces  of  silver  in  a  pound,  or  16  ounces,  of  the 

first  ingot,  it  follows  that  one  ounce  of  it  contains  ^-^  of  an 

ounce  of  silver,  and  consequently  in  a  number  of  ounces 

Ix 
denoted  by  aj,  there  is   -—   ounces  of  silver.      In  the  same 
16 

10* 


114  FIRST  LESSONS  IN  ALGEBRA. 

manner  we  would  find  that   -— ^  and  -— ,  express  the  num- 

Ib  lb 

ber  of  ounces  of  silver  taken  from  the  second  and  third,  to 

form  the  fourth  ;  but  from  the  enunciation,  one  pound  of  this 

fourth  ingot  contains  8  ounces  of  silver.     We  have,  then, 

for  the  first  equation, 

16^    16       16 
or,  making  the  denominators  disappear, 
7a;+12y  +  4^==128. 

As  respects  the  copper,  we  should  find 

3x+37/+7z=z60, 
and  with  reference  to  the  pewter 

6x+y+5z=:68, 

As  the  coefficients  of  y  in  these  three  equations,  are 
the  most  simple,  it  is  most  convenient  to  eliminate  this  un- 
known quantity  first. 

Multiplying  the  second  equation  by  4,  and  subtracting 
the  first,  we  have 

DX-{-2iZz=:ll2. 

Multiplying  the  third  equation  by  3,  and  subtracting  the 
second  from  the  product, 

150^4-80  =  144. 

Multiplying  this  last  equation  by  3,  and  subtracting  the 
preceding  one  from  the  product,  we  obtain 

40a?  =:  320, 

whence  a? =8. 


EQUATIONS    OF    THE    FIRST    DEGREE.  115 

Substitute  this  value  for    x   in  the  equation 

150^+8^=144  ; 

it  becomes  120-f-8^  =  144, 

whence  zz=^3. 

Lastly,  the  two  values  x=8,  z  —  d,  being  substituted  in 
the  equation 

Qx+y  +  ^z  —  QS, 

give  ^  484-y+15=:::68, 

whence  y  =  ^- 

Therefore,  in  order  to  form  a  pound  of  the  fourth  ingot, 
we  must  take  8  ounces  of  the  first,  5  ounces  of  the  second, 
and  3  of  the  third. 

Verification. 

If  there  be  7  ounces  of  silver  in  16  ounces  of  the  first 
ingot,  in  8  ounces  of  it, there  should  be  a  number  of  ounces 
of  silver  expressed  by 

7x8 
.    16  • 
In  like  manner, 

12x5  ,     4x3 

and 


% 


16  16 

will  express  the  quantity  of  silver  contained  in  5  ounces  of 
the  second  ingot,  and  3  ounces  of  the  third. 
Now,  we  have 

7X8      12x5    .  4x3       128       ^ 


16  '  16  '  16  16  ~  ' 
therefore,  a  pound  of  the  fourth  ingot  contains  8  ounces  of 
silver,  as  required  by  the  enunciation.  The  same  condi- 
tions may  be  verified  relative  to  the  copper  and  pewter. 


116  FIRST    LESSONS    IN    ALGEBRA. 

6.  A's  age  is  double  B's,  and  B's  is  triple  of  C's,  and  the 
sum  of  all  their  ages  is  140.     What  is  the  age  of  each  ? 

Ans,  A's  =  84,  B's=:42,  and  C's  =14. 

7.  A  person  bought  a  chaise,  horse,  and  harness,  for 
£60  ;  the  horse  came  to  twice  the  price  of  the  harness, 
and  the  chaise  to  twice  the  price  of  the  horse  and  harness. 
What  did  he  give  for  each  ? 

^  ^13     6s.  Sd.     for  the  horse. 
Ans.    <  jC  6   I3s.  Ad.     for  the  harness. 
'  jC40  for  the  chaise. 

8.  To  divide  the  number  36  into  three  such  parts  that 
i  of  the  first,  i  of  the  second,  and  i  of  the  third,  may  be 
all  equal  to  each  other.  A7is.  8,  12,  and  16. 

9.  If  A  and  B  together  can  do  a  piece  of  work  in  8  days, 
A  and  C  together  in  9  days,  and  B  and  C  in  ten  days  ;  how 
many  days  w^ould  it  take  each  to  perform  the  same  work 
alone  ?  Ans.  A  14f|,  B  17ff,  C  23/i-. 

10.  Three  persons,  A,  B,  and  C,  begin  to  play  together, 
having  among  them  all  $600.  At  the  end  of  the  first  game 
A  has  won  one-half  of  B's  money,  which,  added  to  his  own, 
makes  double  the  amount  B  had  at  first.  In  the  second 
game,  A  loses  and  B  wins  just  as  much  as  C  had  at  the 
beginning,  when  A  leaves  off  with  exactly  what  he  had  at 
first.     How  much  had  each  at  the  beginning  ? 

A71S.  A  $300,  B  $200,  C  $100. 

11.  Three  persons.  A,  B,  and  C,  together  possess  $3640. 
If  B  gives  A  $400  of  his  money,  then  A  will  have  $320 
more  than  B  ;  but  if  B  takes  $140  of  C's  money,  then  B 
and  C  will  have  equal  sums.     How  much  has  each  ? 

Ans.  A  $800,  B  $1280,  C  $1560. 

12.  Three  persons  have  a  bill  to  pay,  which  neither 
alone  is  able  to  discharge.  A  says  to  B,  "  Give  me  the 
4th  of  your  money,  and  then  I  can  pay  the  bill."  B  says 
to  C,  "  Give  me  the  8th  of  yours,  and  I  can  pay  it.     But 


EQUATIONS    OF    THE    FIRST    DEGREE.  117 

C  says  to  A,  "  You  must  give  me  the  half  of  yours  before 
I  can  pay  it,  as  I  have  but  $8."  What  was  the  amount  of 
their  bill,  and  how  much  money  had  A  and  B  ? 

,         j  Amount  of  the  bill,  $13. 
(  A  had  $10,  andB  $12. 

13.  A  person  possessed  a  certain  capital,  which  he  placed 
out  at  a  certain  interest.  Another  person,  who  possessed 
10000  dollars  more  than  the  first,  and  who  put  out  his  capi- 
tal 1  per  cent,  more  advantageously,  had  an  income  greater 
by  800  dollars.  A  third  person,  who  possessed  15000  dol- 
lars more  than  the  first,  putting  out  his  capital  2  per  cent, 
more  advantageously,  had  an  income  greater  by  1 500  dollars. 
Required  the  capitals  of  the  three  persons,  and  the  rates  of 
interest. 

^^^   5  Sums  at  interest,     $30000,     40000,     45000. 


■\ 


Rates  of  interest,  4  5  6  pr.  ct. 

14.  A  widow  receives  an  estate  of  $15000  from  her  de- 
ceased husband,  with  directions  to  divide  it  among  two  sons 
and  three  daughters,  so  that  each  son  may  receive  twice  as 
much  as  each  daughter,  and  she  herself  to  receive  $1000 
more  than  all  the  children  together.  What  was  her  share, 
and  what  the  share  of  each  child  ? 

^  The  widow's  share,  $8000. 

Ans.  <  Each  son's,  2000. 

(  Each  daughter's,  1000. 

15.  A  certain  sum  of  money  is  to  be  divided  between 
three  persons.  A,  B,  and  C.  A  is  to  receive  $3000  less 
than  half  of  it,  B  $1000  less  than  one  third  part,  and  C  to 
receive  $800  more  than  the  fourth  part  of  the  whole.  What 
is  the  sum  to  be  divided,  and  what  does  each  receive  ? 

r  Sum,  $38400. 


Ans.  < 


A  receives     16200. 
B        „  11800. 

L  C        „  10400 


118  FIRST  LESSONS  IN  ALGEBRA. 


CHAPTER  IV. 

Of  Powers, 

19.  If  a  quantity  be  multiplied  several  times  by  itself, 
the  product  is  called  the  power  of  the  quantity.     Thus, 

az=za  is  the  root,  or  first  power  of  a. 

ay^arz^a^  is  the  square,  or  second  power  of  a, 

aXaXa  —  a^  is  the  cube,  or  third  power  of  a. 

aXaXaXa=a^  is  the  fourth  power  of  a. 

axaxaxaxaz=:a^  is  the  fifth  power  of  a. 

In  every  power  there  are  three  things  to  be  considered  • 

1st.  The  quantity  which  is  multiplied  by  itself,  and  which 
is  called  the  root  or  the  first  power. 

2nd.  The  small  figure  which  is  placed  at  the  right,  and 
a  little  above  the  letter.  This  figure  is  called  the  exponent 
of  the  power,  and  shows  how  many  times  the  letter  enters 
as  a  factor. 

3rd.  The  power  itself,  which  is  the  final  product,  or 
result  of  the  multiplications. 

Quest. — 79.  If  a  quantity  be  continually  multiplied  by  itself,  what  is 
the  product  called  1  How  many  things  are  to  be  considered  in  every 
power  1     What  are  thoy ! 


OF   POWERS.  IIP 


For  example,  if  we  suppose  «  =  3,  we  have 

«=     3  the  root,  or  1st  power  of  3. 

«2  — 32--3x3=:     9  the  second  power  of  3. 

a"5  — 33  =  3  X3  X3=:   27  the  third  power  of  3. 

a^rr 3^=3  X3   X3x3=:   81  the  fourth  power  of  3. 

flS  — 35  =  3x3  X3x  3  X3=r243  the  fifth  power  of  3. 

In  these  expressions,  3  is  the  root,  1,  2,  3,  4  and  5  are 
the  exponents,  and  3,  9,  27,  81  and  243  are  the  powers. 


To  raise  monomials  to  any  potoer, 

80.  Let  it  be  required  to  raise  the  monomial  2a?h'^  to 
the  fourth  power.     We  have 

(2a352)4=2«3^,3  y^  2a3J2  X  2^352  X  2a?h-, 

which  merely  expresses  that  the  fourth  power  is  equal  to 
the  product  which  arises  from  writing  the  quantity  four 
times  as  a  factor.  By  the  rules  for  multiplication,  this  pro- 
duct becomes 

(2a3Z>2y4^2'^a3  +  3  +  3  +  3j2  +  2  +  2  +  2_2%12^,8. 

from  which  we  see, 

1st.  That  the  coefficient  2  must  be  raised  to  the  4th 
power ;  and, 

"  2nd.  That  the  exponent  of  each  letter  must  be  multiplied 
by  4,  the  exponent  of  the  power. 

As  the  same  reasoning  would  apply  to  every  example, 
we  have,  for  the  raising  of  monomials  to  any  power,  the 
following 


120  FIRST  LESSONS  IN  ALGEBRA. 

RULE 

I.  Raise  the  coejficient  to  the  required  power, 

II.  Multiply  the  exponent  of  each  letter  hy  the  exponent  of 
the  power. 

EXAMPLES. 

1 .  What  is  the  square  of   Sce^y^  ?  Ans.  9a'^y^. 

2.  What  is  the  cube  of   Qa^y-x  1  Ans.  216«'ya;^. 

3.  W^hat  is  the  fourth  power  of  2d^y%^  1 

4.  What  is  the  square  of  a-¥'y^  ?  Ans.  a^h^^y^. 

5.  What  is  the  seventh  power  of  a%c(P  1 

Ans.  a^-Wc'd?\ 

6.  What  is  the  sixth  power  of  a%^c^d  1     Ans.  a^%^^c^"d^. 

7.  What  is  the  square  and  cube  of    —2a-lr  ? 

Square.  Cube. 

— 2^253  —2a%- 

-2^252  '                                 -2^ 

-^Aa^¥  -\-Aa^¥ 

"  -2a2^3 


By  observing  the  way  in  which  the  powers  are  formed, 
we  may  conckide, 

1st.    When  the  root  is  positive,  all  the  powers  will  he  positive. 
2nd.   When  the  root  is  negative,  all  the  even  powers  tcill  be 
positive  and  all  the  odd  powers  negative. 

Quest. — 80.  What  is  a  monomial  1  Give  the  rule  for  raising  a 
monomial  to  any  power.  When  the  root  is  positive,  how  will  the  powers 
be  1     When  the  root  is  negative,  how  will  the  powers  be  1 


OF    POWERS.  121 

8.  What  is  the  square  of   •—2a^¥  ?  Ans.  Aa^^^, 

9.  What  is  the  cube  of  —^a^y'^cl        A7is,   —\2^a}^y^c'^» 

10.  What  is  the  eighth  power  of   —d?xy^  ? 

Ans.   +a2%8yi6^ 

11.  What  is  the  seventh  power  of    —cP-yx^  ? 

Ans.   — a^^y^x^^, 

12.  What  is  the  sixth  power  of  2a¥y^  1 

Ans.  Qia%'^^y'^^. 

13.  What  is  the  ninth  power  of    — cdx^y"^  ? 

Ans.     —  c9(Z9a;18y27, 

14.  What  is  the  sixth  power  of    —'^ah'^d  ? 

Ans.  729 a^^'^d^ 

15.  What  is  the  square  of    —lOa^^c^  1      Ans.   WOa'^Mc^, 

16.  What  is  the  cube  of    —9a%^d-p  ? 

Ajis.   —129a}%^^d^p. 

17.  What  is  the  fourth  pow-er  of    —Aa^h^c'^d''  1 

Ans.  256a20^,i2ci6^20. 

18.  What  is  the  cube  of    —4:a?hVd  ? 

Ans.   —QAa%^cH^. 

19.  What  is  the  fifth  power  of  2w^lrxy  ? 

Ans.  Z2a}^h^^x^y^. 

20.  What  is  the  square  of  20a:^y^c5  ?        Ans.  AOOx^y^c^^. 

21.  What  is  the  fourth  power  of  ^a-h\^  ? 

Ans.  Sla%^c^'^. 

22.  What  is  the  fifth  power  of  —c^d'^x'^y'^  ? 

Ans.   -ciojiVOyio. 

23.  What  is  the  sixth  power  of  —ac^dfl 

Ans.  a^c^'^d^f^. 

24.  What  is  the  fourth  power  of  —2aP'c'^d^  1 

Ans.  l^a^c^d^^. 
11 


122  FIRST  LESSONS  IN  ALGEBRA. 


To  raise  Polynomials  to  any  power. 

8 1 .  The  power  of  a  polynomial,  like  tliat  of  a  mono- 
mial, is  obtained  by  multiplying  the  quantity  continually  by 
itself.  Thus,  to  find  the  fifth  power  of  the  binomial  a~\-h, 
we  have 

a  -{-  h 1st  power. 

a+  h 

a^-j-  ah 

+  ah+h'^ 


a^-\-2ah  -\-P 2nd  power. 

a  -{-   b 

+   a^b-}-   2ab'^  +    h^ 
a^+da^-i-   3ab^  +   b^    .     .     .     3rd  power. 
a  +   b 

+   a^+  3a^'^-\-3ab^     +  b^ 

a'^-{-ia^-{-  6a^^+   4ab^  +  5^     4th  power. 
a  +   b 

a^-\-4a^b-{-  ea^^+~4^^3 -{-  ab^ 

a^^^a^b+lOa%'^+\OaV)'i+bab^-^b^     Ans. 


Remark. — 82.  It  will  be  observed  that  the  number  of 
multiplications  is  always  1  less  than  the  imits  in  the  expo- 

QuEST. — 81.  How  is  the  power  of  a  polynomial  obtained. 


OF    POWERS 


123 


nent  of  the  power.  Thus,  if  the  exponent  is  1,  no  multipli- 
cation is  necessary.  If  it  is  2,  we  multiply  once  ;  if  it  is 
3,  twice  ;  if  4,  three  times,  &c.  The  powers  of  polyno- 
mials may  be  expressed  by  means  of  the  exponent.  Thus, 
to  express  that  a-^-b  is  to  be  raised  to  the  5th  power,  we  write 

(a+bY, 

which  expresses  the  fifth  power  of  a+b. 

2.  Find  the  5th  power  of  the  binomial  a—b. 


a  ~   b 

*""  r""^** 

a^ —  ab 

—   ab  +62 

a^-2ab  +5^     . 

2nd  power. 

a  -     b 

a^-2a^b^ 

ah"^ 

-   a%^ 

2«52  _ 

-53 

a?—3a%^ 

3a&2  _ 

-53      ...     3rd  power. 

a  -   b 

a^-Sa%  + 

3a2^,2_ 

-     ab^ 

-   a?b  + 

3^252- 

-   3ab^  +   b^ 

a^—A:a?b  + 

6^2^.2  _ 

-  Aah^  +   ¥     4th  power. 

a  ~   b 

a^-Aa^b  + 

QaW- 

-   4^253+   ab^ 

—   a^b  + 

4a^^- 

-   Qa%^-^4:ab^-b^ 

^5  _  5^45  _|_  10^352 - 

-I0a^^+5ab^-b^     Ans, 

Quest. — 82.  How  does  the  number  of  multiplications  compare  with 
the  exponent  of  the  power  1  If  the  exponent  is  4,  how  many  multipli- 
cations 1 


124  FIRST    LESSONS    IN    ALGEBRA. 

3.  What  is  the  square  of    5a—2c  +  d, 

5a  —  2c  +     d 

5a  —  2c  -\-     d 

25«2_l0ac+   5ad 

—  lOac-i-   4c^  —2cd 

+   5ad—2cd+   d^ 
25a^-20ac+l0ad-\-Ac'^  —A^cd+d^     Ans. 

4.  Find  the  4th  power  of  the  binomial     3a— 25. 

^a  —  2h 1st  power. 

3a  —  2h 

9^2  _  Qah 

—  Qah  +  Ah'^ 

9a^—  l2ah-\-Ah'^ 2nd  power. 

3«  —  25 


27«3-,   36^25+ 12a52 

27a3_  54a25+36«52—     8^3    .     .     3rd  power. 

3a  —     2h 

81a^-162a354-108a252_24«J3 

■^   5Aa'^h-\-\0Sa%'^—l[2a¥ -{-!&¥ 
Sla^— 216a3Z> -1-216^252  — 96a53-f  165^     Ans. 

5.  What  is  the  square  of  the  binomial  a+1 1 

Ans.  a2_j_2«-|_l. 

6.  What  is  the  square  of  the  binomial  a  — 11 

Ans.  a^—2a-\-l, 

7.  What  is  the  cube  of  9a  — 35  ? 

Ans,  729a3— 729a25+243a52_2753, 

8.  What  is  the  third  power  of  a— 1  ? 

Ans.  a3_3^2_|_3^_l 


OF    POWERS.  125 

9.  What  is  the  4th  power  of  x—yl 

Arts,  a?*— 4a;3y+6a;y— 4a7y3  +  y*. 

10.  What  is  the  cube  of  the  trinomial  ipc+y+zl 

Ans.  x^  +  Zxhj+2x'^z-\-'^xy'^+^xz'^+'^i/z-\-Zyz'^+Qxyz 
-{-y^-\-z^. 

11.  What  is  the  cube  of  the  trinomial  2a'^—'4ab-\-3b^  1 
Ans.  Sa^—4Sa^b+l32a'^b'^—208a^^+l9Sa^b^—l0Sab^ 

+  21¥. 

To  7^aise  a  Fraction  to  any  Poioer. 

83.  The  power  of  a  fraction  is  obtained  by  multiplying 
the  fraction  by  itself ;  that  is,  by  multiplying  the  numerator 
by  the  numerator,  and  the  denominator  by  the  denominator. 

Thus,  the  cube  of  ^-,    which  is  written 
b 


/  a  \^      a        a       a        a^ 


is  found  by  cubing  the  numerator  and  denominator  sepa- 
rately. 

2.  What  is  the  square  of  the  fraction     7- —  ? 

b-{-c 

We  have 

/a  —  c  \2^  {a—cY  _  a?'—2ac-\-c^ 
\b^)  ~  [b+cf  ~"P+2^+^       ^^' 

3.  What  is  the  cube  of    ^-  1  Ans.    -~^. 

3bc  27^>V 


QuEST.^-83.    How  do  you  find  the  power  of  a  fraction  1 
11* 


126  FIRST  LESSONS  IN  ALGEBRA. 


4.  What  is  the  fourth  power  of    — r—  1 


Ans. 


1  Qxhj^  ' 


5.  What  is  the  cube  of    - — ^  ? 

x-{-y 

x"^ — ^x^y + 3a?y2 — y^ 

x'^-\-^x'^y-\-'ixy'^-\-y^* 

Q>ctx  '^'^ 

6.  What  is  the  fourth  power  of    ?         Ans. 


Aay  1  Qy"^ 

7.  What  is  the  fifth  power  of  ?        ^;?^. 


8.  What  is  the  square  of 


IQyz  32y^z^ 

ax—y^ 


by  —  a: 


dP'x'^  —  2axy-\-y'^ 
^y — '^hxy-\-x^ 


9.  What  is  the  cube  of    -^—- —  ? 
x-\-2y 


x'^-\-Qx'^y-\-\2xy'^-\-Sy'^ 


Binomial  Theorem. 

8  4.  The  method  which  has  been  explained  of  raising  a 
polynomial  to  any  power,  is  somewhat  tedious,  and  hence 
other  methods,  less  difficult,  have  been  anxiously  sought 
for.  The  most  simple  which  has  yet  been  discovered,  is 
the  one  invented  by  Sir  Isaac  Newton,  called  the  Binomial 
Theorem. 


Quest. — 84.    What  is  the  object  of  the  Binomial  Theorem  1     Who 
discovered  this  theorem  \ 


BINOMIAL    THEOREM.  127 

85.  In  raising  a  quantity  to  any  power,  it  is  plain  that 
there  are  four  things  to  be  considered  : — 

1st.  The  number  of  terms  of  the  power. 
2nd.  The  signs  of  the  terms. 
3rd.  The  exponents  of  the  letters. 
4th.  The  coefficients  of  the  terms. 

Of  the  Terms. 

86.  If  we  take  the  two  examples  of  Article  81,  which 
we  there  wrought  out  in  full ;  we  have 

(^+^,)5,^^5_L.5«4^_|_10a3^,3_|_10«253_{_5^j4_^55   . 

By  examining  the  several  multiplications,  in  Art.  8 1 ,  we 
shall  observe  that  the  second  power  of  a  binomial  contains 
three  terms,  the  third  power  four,  the  fourth  power  five,  the 
fifth  power  six,  &c  ;  and  hence  we  may  conclude — That 
the  number  of  terms  in  any  power  of  a  binomial,  is  one  greater 
than  the  exponent  of  the  power. 

Of  the  Signs  of  the  Terms. 

87.  It  is  evident  that  when  both  terms  of  the  given  bi- 
nomial are  plus,  all  the  terms  of  the  poicer  will  be  plus. 

2nd.  If  the  second  term  of  the  binomial  is  negative,  then 
all  the  odd  terms,  counted  from  the  left,  ivill  be  positive,  and 
all  the  even  terms  necrative. 


Quest. — 85.  In  raising  a  quantity  to  any  power,  how  many  things 
are  to  be  considered'?  What  are  theyl — 86.  How  many  terms  are 
there  in  any  power  of  a  binomial  1  If  the  exponent  is  3,  how  many 
terms'?  If  it  is  4,  how  many  terms'?  If  5'?  &c. — 87.  If  both  terms 
of  the  binomial  are  positive,  how  are  the  terms  of  the  power '?  If  the 
second  term  is  negative,  how  are  the  signs  of  the  terms  1 


128  FIRST  LESSONS  IN  ALGEBRA. 

Of  the  Exponents. 

88.  The  letter  which  occupies  the  first  place  in  a  bino- 
mial, is  called  the  leading  letter.  Thus,  a  is  the  leading 
letter  in  the  binomials    a-\-h,    a—h. 

1st.  It  is  evident  that  the  exponent  of  the  leading  letter 
in  the  first  term  will  be  the  same  as  the  exponent  of  the 
power  ;  and  that  this  exponent  will  diminish  by  unity  in 
each  term  to  the  right,  until  we  reach  the  last  term,  which 
does  not  contain  the  leading  letter. 

2nd.  The  exponent  of  the  second  letter  is  1  in  the  second 
term,  and  increases  by  unity  in  each  term  to  the  right, 
until  we  reach  the  last  term,  in  which  the  exponent  is  the 
same  as  that  of  the  given  power. 

3rd.  The  sum  of  the  exponents  of  the  two  letters,  in  any 
term,  is  equal  to  the  exponent  of  the  given  power.  This 
last  remark  will  enable,  us  to  verify  any  result  obtained  by 
the  binomial  theorem. 

Let  us  now  apply  these  principles  in  the  two  following 
examples,  in  which  the  coefficients  are  omitted : — 

{a+hf  .  .  .  a^+a^b-{-a^h'^-^a?P  +  a^¥  +  ah^-{-h\ 
{a  —  bf,..  a^—a^h+a^h'^—a?P+a?¥—ah^-\-h^. 

As  the  pupil  should  be  practised  in  writing  the  terms  with- 
out the  coefficients  and  signs,  before  finding  the  coefficients, 
we  will  add  a  few  niore  examples. 


Quest. — 88.  Which  is  the  leading  letter  of  the  binomial  1  What  is 
the  exponent  of  this  letter  in  the  first  term  1  How  does  it  change  in  the 
terms  towards  the  right  1  What  is  the  exponent  of  the  second  letter  in 
the  second  term  '\  How  does  it  change  in  the  terms  towards  the  right  1 
What  is  it  in  the  last  term  1  What  is  the  sum  of  the  exponents  in  any 
term  equal  to*? 


BINOMIAL    THEOREM.  129 

1.  {a-^-hy  .  .  a^  +  aH-\-ab^^P. 

2.  (a— by  .  .  a'^—a^b  +  a'^b'^  —  ab^+hK 

3.  {a+by  .  .  a^+a^b+a^^+a^^+ab^+b^. 

4.  {a-by  .  .  a^-a%+a^b^-a^b^  +  a^^-a'^b^+ab^-b'^. 

Of  the  Coefficients. 

89.  The  coefficient  of  the  first  term  is  unity.  The  co- 
efficient of  the  second  term  is  the  same  as  the  exponent  of 
the  given  power.  The  coefficient  of  the  third  term  is  found 
by  multiplying  the  coefficient  of  the  second  term  by  the 
exponent  of  the  leading  letter,  and  dividing  the  product  •♦y 
2.  And  finally — If  the  coefficient  of  any  term  be  multiplied 
by  the  exponent  of  the  leading  letter^  and  the  product  divided 
hy  the  number  which  marks  the  place  of  that  term  from  the 
left,  the  quotient  will  be  the  coefficient  of  the  next  term. 

Thus,  to  find  the  coefficients  in  the  example 

{a-by   .  .  ,  a''-a^b  +  a^b^-a^P-\-a^^-a^'+ab'-h'' 

we  first  place  the  exponent  7  as  a  coefficient  of  the  second 
term.  Then,  to  find  the  coefficient  of  the  third  term,  we 
multiply  7  by  6,  the  exponent  of  a,  and  divide  by  2.  The 
quotient  21  is  the  coefficient  of  the  third  term.  To  find  the 
coefficient  of  the  fourth,  we  multiply  21  by  5,  and  divide 
the  product  by  3  :  this  gives  35.  To  find  the  coeffi.cient  of 
the  fifth  term,  we  multiply  35  by  4,  and  divide  the  product 
by  4  :  this  gives  35.  The  coefficient  of  the  sixth  term, 
found  in  the  same  way,  is  21  ;  that  of  the  seventh,  7  ;  and 
that  of  the  eighth,  1.  Collecting  these  coefficients,  we 
have 

(a-by  = 
«7_7a6j_2io6j2_35a4j3+35a3j4_2ia3J6+7a5«-i'. 


130  FIRST  LESSONS  IN  ALGEBRA. 

Remark. — We  see,  in  examining  this  last  result,  tliat  the 
coefficients  of  the  extreme  terms  are  each  unity,  and  that 
the  coefficients  of  terms  equally  distant  from  the  extreme 
terms  are  equal.  It  will,  therefore,  be  sufficient  to  find  the 
coefficients  of  the  first  half  of  the  terms,  from  which  the 
others  may  be  immediately  written. 

EXAMPLES. 

1.  Find  the  fourth  power  of  a-{-h. 

2.  Find  the  fourth  power  of  a  —  b. 

Ans.   a^—4a^+6a'^b^  —  Aab^-{-b\ 

3.  Find  the  fifth  power  of  a+b. 

Ans.  a^-i-5a'^b-\-10a^'^-i-l0a'^b^-\-5ab^-{-b\ 

4.  Find  the  fifth  power  of  a  —  b. 

A?is.   a^  —  ba^b+lOaW—lOa^^-i-^ab^—b^. 

5.  Find  the  sixth  power  of  a-\-b. 

Ans.  a^-i-6a^b-\-l5a'^b'^+20a^^-\-l5a'^b^+6a¥  +  b^ 

6.  Find  the  sixth  power  of  a  —  b. 

Ans.  a^  —  6a^b+l  ba^b^ — 20aW  + 1  ba^"^ —6ab^-\-  ¥. 

7.  Let  it  be  required  to  raise  the  binomial  ^a'^c—2bd  to 
the  fourth  power. 

It  frequently  occurs  that  the  terms  of  the  binomial  are 
affected  with  coefficients  and  exponents,  as  in  the  above 


Quest. — 89.  What  is  the  coefficient  of  the  first  term'!  What  is  the 
coefficient  of  the  second  1  How  do  you  find  the  coefficient  of  the  third 
term  %  How  do  you  find  the  coefficient  of  any  term  1  What  are  the 
coefficients  of  the  first  and  last  terms  1  How  are  the  coefficients  of 
terms  equally  distant  from  the  extremes  1 


BINOMIAL    THEOREM.  131 

example.     In  the  first  place,  we  represent  each  term^of  the 
binomial  by  a  single  letter.     Thus,  we  place 

3a^c=x,     and     ~2bd='i/, 
we  then  have 

{jc + y  )^ = ^'^ + 4aj^y  +  6a;2y2  _|_  4ajy3  _j_  y4^ 

But,  a;2  — 9a4c3,     x^::^27a^c^,     x^  =  Sla^c^ ; 

and  f  ==  ib-^d^,  y^  =  -  Sb^d^,     1/  =  1 6b^d\ 

Substituting  for    x    and    y    their  values,  we  have 

{3a^c-27jd)^=(3a^-cy-\-4{3a^cy{-2bd)  +  6  {3a^cY  {-2bdY 
+  A{3a^c)  {-2bdf+[-2bd)\ 

and  by  performing  the  operations  indicated, 
{3ci^c-2bdY=zS\a^c^-2lQa^c^d  +  2lQa^c%'^d'^-9Qa?c¥d^ 
-^\Q¥d\ 

8.  What  is  the  square  of    3a  —  Qb1 

Arts.  9a2_36^5_j_36^2 

9.  What  is  the  cube  of     3a: — 6y  ? 

Ans.     27a:3— 162a:2y  +  324a:y2_21653. 

10.  What  is  the  square  of     x—y  ? 

Ans.  x^" — 2x1/ -\-y^. 

11.  What  is  the  eighth  power  of  7n-{-n  ? 

A71S.    7n^+8jn/n-{-2Sm^n%56m^n^-{-  70772%^+ 5 6 w^s. 
+  28m^n^+8?nn^-\-n^, 

12.  W^hat  is  the  fourth  power  of     a~3b  1 

Ans.  a^  —  1 2^35 + 54^2^2  __  i  Qgab^ + 8 1 R 

13.  What  is  the  fifth  power  of     c—2d  1 

Ans.  c^  —  \0cH^A0cH'^  —  Q0cH^  +  SQcd^-^32d^. 

14.  What  is  the  cube  of     ba  —  3d  1 

Ans.   125^3  >-.225aV+135a(/2__27rf3 


132  FIRST  LESSONS  IN  ALGEBRA. 

Remark.  The  powers  of  any  polynomial  may  easily  be 
found  by  the  Binomial  Theorem. 

15.  For  example,  raise    a-\-b-\-c    to  the  third  power. 
First,  put    ....    h-\-c=:d. 

Then,  -  {a+h-\-cYz={a-\-dy  =  a^-\-3a^d-{-3ad^+d\ 
Or,  by  substituting  for  the  value  of   d, 

3a^c  -{-3b^c-{-  6abc 

-\-3ac"-{-3bc'^ 

+     c^. 

This  expression  is  composed  of  the  cubes  of  the  three 
terms,  plus  three  times  the  square  of  each  term  by  the  first 
powers  of  the  two  others,  plus  six  times  the  product  of  all 
three  terms.  It  is  easily  proved  that  this  law  is  true  for  any 
polynomial. 

To  apply  the  preceding  formula  to  the  development  of 
the  cube  of  a  trinomial,  in  which  the  terms  are  affected 
with  coefficients  and  exponents,  designate  each  term  by  a 
single  letter,  then  replace  the  letters  introduced,  by  their  values, 
and  perform  the  operations  indicated. 

From  this  rule,  we  find  that 

{2o?'—Aab^3b'^fz:^Sa^—AQa^b-{-\32a^b'^—20Sa^b^ 
+  198a2Z,4_i08a65+2756. 

The  fourth,  fifth,  &c,  powers  of  any  polynomial  can  be 
found  in  a  similar  manner. 

16.  What  is  the  cube  of     a— 2Z>-f  c  ? 

Ans.  a3_853+c3  — 6ft2^,-f  3a2c+12a&2+12^2c+3ac2 
—  ebc^-—l2abc. 


^ 


EXTRACT  mv  OF  THE  SQUARE  ROOT.       1 33 


CHAPTER   V. 

Extra*:non  of  the  Square  Root  of  Numbers.  Forma- 
tion of  the  Square  and  Extraction  of  the  Square 
Root  of  Algebraic  Quantities.  Calculus  of  Radicals 
of  the  Second.  Degree. 

90.  The  square  or  second  power  of  a  number,  is  the 
product  which  arises  from  multiplying  that  number  by  itself 
once  :  for  example,  49  is  the  square  of  7,  and  144  is  the 
square  of  12. 

91.  The  square  root  of  a  number  is  that  number  which, 
being  multiplied  by  itself  once,  will  produce  the  given  num- 
ber. Thus,  7  is  the  square  root  of  49,  and  12  the  square 
root  of  144:    for,  7x7=^49,  and  12x12=:  144. 

92.  The  square  of  a  number,  either  entire  or  fractional, 
is  easily  found,  being  always  obtained  by  multiplying  this 
number  by  itself  once.  The  extraction  of  the  square  root 
of  a  number  is,  however,  attended  with  some  difficulty,  and 
requires  particular  explanation. 


Quest. — 90.    What  is  the  square,  or  second  power  of  a  number! — 
91.    What  is  the  square  root  of  a  number  1 

12 


1 


134  FIRST  LESSONS  IN  ALGEBRA* 

The  first  ten  numbers  are. 

1,     2,     3,       4,       5,       6,       7,       8,       9,        10; 

and  their  squares, 

1,     4,     9,     16.     25,     36,     49,     64,     81,     100; 

and  reciprocally,  the  numbers  of  the  first  line  are  the  square 
roots  of  the  corresponding  niunbers  of  the  second.  We 
may  also  remark  that,  the  square  of  a  number  expressed  hy  a 
single  figure^  will  contain  no  figure  of  a  higher  denomination 
than  tens. 

The  numbers  of  the  last  line,  1,  4,  9,  16,  &c,  and  all 
other  numbers  which  can  be  produced  by  the  multiplication 
of  a  number  by  itself,  are  called  perfect  squares. 

It  is  obvious  that  there  are  but  nine  perfect  squares  among 
all  the  numbers  which  can  be  expressed  by  one  or  tAVO 
figures  :  the  square  roots  of  all  other  numbers  expressed 
by  one  or  two  figures,  will  be  found  between  two  whole 
numbers  differing  from  each  other  by  unity.  Thus  55, 
which  is  comprised  between  49  and  64,  has  for  its  square 
root  a  number  between  7  and  8.  Also  91,  which  is  com 
prised  between  81  and  100,  has  for  its  square  root  a  number 
between  9  and  10. 

03.  Every  number  may  be  regarded  as  made  up  of  a 
certain  number  of  tens  and  a  certain  number  of  units. 
Thus  64  is  made  up  of  6  tens  and  4  units,  and  may  be  ex 
pressed  under  the  form  60  +  4. 


Quest. — 92.  What  will  be  the  highest  denomination  of  the  square 
of  a  number  expressed  by  a  single  figure  '?  What  are  perfect^  squares  1 
How  many  are  there  between  1  and  100  1     What  are  they  ? 


EXTRACTION  OF  THE  SQUARE  ROOT.      135 

Now,  if  we  represent  the  tens  by  a  and  the  units  by  h, 
we  shall  have 

a^h    =64, 
and  (a+^)^  =  (64)2; 

or  a^+2ah  +  h'^zzzA09Q. 

Which  proves  that  the  square  of  a  number  composed  of 
tens  and  units,  contains  the  square  of  the  tens  plus  twice  the 
product  of  the  tens  hy  the  units,  plus  the  square  of  the  units. 

94.  If,  now,  we  make  the  units  1,  2,  3,  4,  &c,  tens,  or 
units  of  the  second  order,  by  annexing  to  each  figure  a  ci- 
pher, we  shall  have 

10,    20,    30,      40,       50,       60,       70,       80,       90,       100, 

and  for  their  squares, 

100,  400,  900,  1600,  2500,  3600,  4900,  6400,  8100,  10000. 

From  which  w^e  see  that  the  square  of  one  ten  is  100,  the 
square  of  two  tens  400  ;  and  generally,  that  the  square  of 
tens  will  contain  no  figures  of  a  less  denomination  than  hun- 
dreds, nor  of  a  higher  name  than  thousands. 

Ex.  1. — To  extract  the  square  root  of  6084. 

Since  this  number  is  composed  of  more  than 
two  places  of  figures,  its  roots  wdll  contain  60  84 

more  than  one.     But  since  it  is  less  than  10000, 
which  is  the  square  of  100,  the  root  will  contain  but  two 
figures  :  that  is,  units  and  tens. 

Now,  the  square  of  the  tens  must  be  found  in  the  two 


Quest. — 93.  How  may  every  number  be  regarded  as  made  up  1  What 
is  the  square  of  a  number  composed  of  tens  and  units  equal  to? — ■ 
94.  What  is  the  square  of  one  ten  equal  to  1  Of  2  tens  1  Of  3 
tens'!  &.C. 


60  84 

49 

1184 

1184 

136  FIRST  LESSONS  IN  ALGEBRA. 

left-hand  figures,  wliicli  v/e  will  separate  from  the  other  two 
by  putting  a  point  over  the  place  of  units,  and  a  second  over, 
the  place  of  hundreds.  These  parts,  of  two  figures  each,  are 
called  periods.  The  part  60  is  comprised  between  the  two 
squares  49  and  64,  of  which  the  roots  are  7  and  8  :  hence, 
7  is  the  figure  of  the  tens  sought ;  and  the  required  root  is 
composed  of  7  tens  and  a  certain  number  of  units. 

The  figure  7  being  found,  we 
write  it  on  the  right  of  the  given  60  84    78 

number,  from  which  we  separate 
it  by  a  vertical  line:   then  we       7x2  =  141 
subtract  its  square,  49,  from  60, 
which  leaves  a  remainder  of  1 1 ,  q~ 

to  which  we  bring  down  the  two 

next  figures  84.  The  result  of  this  operation,  1184,  con- 
tains twice  the  product  of  the  tens  hy  the  units,  plusUhe  square 
of  the  units. 

But  since  tens  multiplied  by  units  cannot  give  a  product  of 
a  less  name  than  tens,  it  follows  that  the  last  figure,  4,  can 
form  no  part  of  the  double  product  of  the  tens  by  the  units : 
this  double  product  is  therefore  found  in  the  part  118,  which 
we  separate  from  the  units'  place,  4. 

Now  if  we  double  the  tens,  which  gives  14,  and  then  di- 
vide 118  by  14,  the  quotient  8  is  the  figure  of  the  units,  or 
a  figure  greater  than  the  units.  This  quotient  figure  can 
never  be  too  small,  since  the  part  118  will  be  at  least  equal 
to  twice  the  product  of  the  tens  by  the  units  :  but  it  may  be 
too  large  ;  for  the  118,  beside-s  the  double  product  of  the 
tens  by  the  units,  may  likewise  contain  tens  arising  from 
the  square  of  the  units.  To  ascertain  if  the  quotient  8  ex- 
presses the  units,  we  write  the  8  on  the  right  of  the  14, 
which  gives  148,  and  then  we  multiply  148  by  8.  Thus, 
we  evidently  form,  1st,  the  square  of  the  units;  and, 
2nd,  the  double  product  of  the  tens  by  the  units.     This 


EXTRACTION  OF  THE  SQUARE  ROOT.       137 

multiplication  being  effected,  gives  for  a  product  1184,  a 
number  equal  to  the  result  of  the  first  operation.  Having 
subtracted  the  product,  we  find  the  remainder  equal  to  0  : 
hence  78  is  the  root  required. 

Indeed,  in  the  operations,  we  have  merely  subtracted 
from  the  given  number  6084,  1st,  the  square  of  7  tens,  or 
70  ;  2nd,  twice  the  product  of  70  by  8  ;  and,  3d,  the  square 
of  8  :  that  is,  the  three  parts  which  enter  into  the  composi- 
tion of  the  square  70-4-8,  or  78  ;  and  since  the  result  of 
the  subtraction  is  0,  it  follows  that  78  is  the  square  root  of 
6084. 

05.  Remark. — The  operations  in  the  last  example  have 
been  performed  on  but  two  periods,  but  it  is  plain  that  the 
same  reasoning  is  equally  applicable  to  larger  numbers,  for 
by  changing  the  order  of  the  units,  we  do  not  change  the 
relation  in  which  they  stand  to  each  other. 

Thus,  in  the  number  60  84  95,  the  two  periods  60  84 
have  the  same  relation  to  each  other  as  in  the  number 
60  84  ;  and  hence  the  methods  used  in  the  last  example 
are  equally  applicable  to  larger  numbers. 

96.  Hence,  for  the  extraction  of  the  square  root  of 
numbers,  we  have  the  following 

RULE. 

I.  Separate  the  given  number  into  periods  of  two  figures 
each,  beginning  at  the  right  hand : — the  period  on  the  left  will 
often  contain  but  one  figure. 

II.  Find  the  greatest  square  in  the  first  period  on  the  left, 
and  place  its  root  on  the  right,  after  the  manner  of  a  quotient 


Quest. — 95.    Will  the  reasoning  in  the  example  apply  to  more  than 
two  periods  1 

12* 


138  FIRST  LESSONS  IN  ALGEBRA. 

m  division.  Subtract  the  square  of  the  root  from  the  first 
period,  and  to  the  remainder  bring  down  the  second  period  for 
a  dividend. 

III.  Double  the  root  already  found,  and  place  it  on  the  left 
for  a  divisor.  Seek  how  many  times  the  divisor  is  contained 
in  the  dividend,  exclusive  of  the  right-hand  figure,  and  place 
the  figure  in  the  root  and  also  at  tlie  right  of  the  divisor. 

IV.  Multiply  the  divisor,  thus  augmented,  by  the  last  figure 
of  the  root,  and  subtract  the  product  from  the  dividend,  and  to 
the  remainder  bring  doicn  the  next  period  for  a  new  dividend. 
But  if  any  of  the  products  should  be  greater  than  the  dividend, 
diminish  the  last  figure  of  the  root. 

V.  Double  the  whole  root  already  found,  for  a  new  divisor, 
and  continue  the  operation  as  before,  until  all  the  periods  are 
brought  down. 

07.  1st  Remark.  If,  after  all  the  periods  are  brought 
down,  there  is  no  remainder,  the  proposed  number  is  a  per- 
fect square.  But  if  there  is  a  remainder,  you  have  only 
found  the  root  of  the  greatest  perfect  square  contained  in 
the  given  number,  or  the  entire  part  of  the  root  sought. 

For  example,  if  it  were  required  to  extract  the  square 
root  of  665,  we  should  find  25  for  the  entire  part  of  the 
root,  and  a  remainder  of  40,  which  shows  that  665  is  not 
a  perfect  square.  But  is  the  square  of  25  the  greatest  per- 
fect square  contained  in  665  1  that  is,  is  25  the  entire  part 
of  the  root  ?  To  prove  this,  we  will  first  show  that,  the 
difference  between  the  squares  of  two  consecutive  numbers,  is 
equal  to  twice  the  less  number  augmented  by  unity. 


Quest. — 96.  Give  the  rule  for  extracting  the  square  root  of  numbers. 

What  is  the  first  step  1     What  the  second  1     What  the  third  1     What 
the  fourth  1     What  the  fifth  1 


EXTRACTION  OF  THE  SQUARE  ROOT.       139 

Let  .  .  az=z    the  less  number, 

and  .  .  a-j-l    =    the  greater. 

Then  .  {a-^\f  —  d^^+2a-\-\, 

and  .  .        [af-z^d^. 

Then*  difference  is       n^        2a+l     as  enunciated. 

Hence,  the  entire  part  of  the  root  cannot  be  augmented, 
unless  the  remainder  exceeds  twice  the  root  found,  plus 
unity. 

But  25  x2+l=:51>40  the  remainder:  therefore,  25  is 
the  entire  part  of  the  root. 

98.  2nd  Rejmark. — The  number  of  figures  in  the  root 
will  always  be  equal  to  the  number  of  periods  into  which 
the  given  number  is  separated. 

EXAMPLES. 

1.  To  find  the  square  root  of  7225.  Ans,  85. 

2.  To  find  the  square  root  of  17689.  Ans.   133. 

3.  To  find  the  square  root  of  994009.  Ans.  997. 

4.  To  find  the  square  root  of  85673536.  Ans.  9256. 

5.  To  find  the  square  root  of  67798756.  Ans.  8234. 

6.  To  find  the  square  root  of  978121.  Ans.  989. 

7.  To  find  the  square  root  of  956484.  Ans.  978. 

8.  What  is  the  square  root  of  3^372961  ?  Ans.  6031. 

9.  What  is  the  square  root  of  22071204  ?  Ans.  4698. 

10.  What  is  the  square  root  of  106929  ?  Ans.  327. 

11.  What  is  the  square  root  of  12088868379025  ? 

Ans.  3476905. 


Quest. — 98.  How  many  %ures  will  you  alwajrs  find  in  the  roott 


140 


FIRST    LESSONS    IN    ALGEBRA. 


99.  3rd  Remark. — If  the  given  number  has  not  an  exac^ 
root,  there  will  be  a  remainder  after  all  the  periods  are 
brought  down,  in  which  case  ciphers  may  be  annexed, 
forming  new  periods,  each  of  which  will  give  one  decimal 
place  in  the  root. 

1.  What  is  the  square  root  of  36729  ? 


In  this  example  there  are 
two  periods  of  decimals, 
which  give  two  places  of 
decimals  in  the  root. 


3  67  29 
1 


191,64  +  . 


2  9  267 
261 

38 

1 

629 
381 

382  6 
3832  ^ 

24800 
22956 

i  184400 
153296 

31104 

Rem. 

2.  What  is  the  square  root  of  2268741  ? 


Ans,  1506,23  +  . 


3.  What  is  the  square  root  of  7596796  ' 


4.  What  is  the  square  root  of  96  ? 


5.  What  is  the  square  root  of  153  ? 


6.  What  is  the  square  root  of  101  ? 


Ans.  2756,22  +  . 


Ans.  9,79795  +  . 


Ans.  12,36931  +  . 


Ans.  10,04987+. 


Quest. — 99.  How  will  you  find  the  decimal  part  of  the  root  1 


EXTRACTION    OF    THE    SQUARE    ROOT.  141 

7.  What  is  the  square  root  of  285970396644  1 

Ans.  534762. 

8.  What  is  the  square  root  of  41605800625  ? 

Ans.  203975. 

9.  What  is  the  square  root  of  48303584206084  ? 

Ans.  6950078. 

Extraction  of  the  square  root  of  Fractions. 

1 OO.  Since  the  square  or  second  power  of  a  fraction  is 
obtained  by  squaring  the  numerator  and  denominator  sepa- 
rately, it  follows  that  the  square  root  of  a  fraction  will  be 
equal  to  the.  square  root  of  the  numerator  divided  by  the 
square  root  of  the  denominator. 

For  example,  the  square  root  of   —   is  equal  to    —  :    for 
a         a        a^ 

1.  What  is  the  square  root  of    —  ? 

9 

2.  What  is  the  square  root  of    —  ? 

64 

3.  What  is  the  square  root  of    ~-  1 

81 

256 

4.  What  is  the  square  root  of 


361 

5.  What  is  the  square  root  of    -—  ? 

64 


Ans. 

1 
2' 

Ans. 

3 

Ans. 

8 

Ans. 

16 
19' 

1 

Ans. 

2' 

Quest. — 100.  If  the  numerator -and  denominator  of  a  fraction  are 

perfect  squares,  how  will  you  extract  the  square  root  T 


142  FIRST  LESSONS  IN  ALGEBRA. 

4096  64 

6.  What  is  the  square  root  of     ^777x7;   ?         ^n^-    ttttt*- 

«    ^Trt.      •     !_  r     582169  ,  ,  763 

7.  Wnat  IS  the  square  root  oi  ?         Ans.    •^^' 

101.  If  neither  the  numerator  nor  the  denominator  is  a 
perfect  square,  the  root  of  the  fraction  cannot  be  exactly 
found.  We  can,  however,  easily  find  the  approximate  root. 
For  this  purpose  ^ 

Multiply/  ho^h  terms  of  the  fraction  hy  the  denominator, 
which  makes  the  denominator  a  perfect  square  without  altering 
the  value  of  the  fraction.  Then,  extract  the  square  root  of 
the  numerator,  and  divide  this  root  hy  the  root  of  the  denomi^ 
tor ;   this  quotient  will  he  the  approximate  root. 

3 

Thus,  if  it  be  required  to  extract  the  square  root  of  — , 

]  5 
we  multiply  both  terms  by  5,  which  gives    — -. 

We  then  have 

-v/TFnr  3,8729+  : 

hence,  3,8729+  -^  5  =:  ,7745+  —  Ans, 

7 

2.  What  is  the  square  root  of    —  ?        Ans.  1,32287  +  . 

14 

3.  What  is  the  square  root  of    -—  ?        A?ts.  1,24721  +  . 


n 

4.  What  is  the  square  root  of     1 1 77:  ? 

16 


Ans.  3,41869+. 


Quest. — IGl.  If  the  numerator  and  denominator  of  a  fraction  are  net 
perfect  squares,  how  do  you  extract  the  square  root  \ 


EXTRACTION  OF  THE  SQUARE  ROOT.      143 

13 

5.  What  is  the  square  root  of     7--  ?      Ans.  2,71313  +  . 

36 

15 

6.  What  is  the  square  root  of     8—  ?      Ans,  2,88203+. 

5 

7.  What  is  the  square  root  of    —  ?        Ans.  0,64549+. 

3 

8.  W^hat  is  the  square  root  of     10—  ? 

Ans.  3,20936+. 

10!3.  Finally,  instead  of  the  last  method,  we  may,  if  we 
please. 

Change  the  vulgar  fraction  into  a  decimal,  and  continue  the 
division  until  the  number  of  decimal  places  is  double  the  number 
of  places  required  in  the  root.  Then,  extract  the  root  of  the 
decimal  by  the  last  rule, 

Ex.  1.  Extract  the  square  root  of    —-     to  within    ,001. 

This  number,  reduced  to  decimals,  is   0.785714  to  within 
0,000001  ;  but  the  root  of  0,785714  to  the  nearest  unit,  is 

.886  ;  hence  0,886  is  the  root  of    —     to  within  ,001. 

14 

2.  Find  the    \/2—     to  within  0,0001. 

V      15 

Ans.  1,6931  +  . 

3.  What  is  the  square  root  of    —  ?        Ans.  0,24253  +  . 

7 

4.  What  is  the  square  root  of    -—  ?        Ans.  0,93541  +  . 

8 

5 

5.  What  is  the  square  root  of    — -  ?        Ans,  1,29099  +  . 

o 


Quest. — 102.  By  what  other  method  may  the  root  be  foimdl 


144  FIRST  LESSONS  IN  ALGEBRA. 

Extraction  of  the  Square  Root  of  Monomials. 

103.  In  order  to  discover  the  process  for  extracting  the 
square  root,  we  must  see  how  the  square  of  the  monomial 
is  formed. 

By  the  rule  for  the  multiplication  of  monomials  (Art.  35), 
we  have 

that  is,  in  order  to  square  a  monomial,  it  is  necessary  to 
square  its  coefficient,  and  double  each  of  the  exponents  of  the 
different  letters.  Hence,  to  fmd  the  root  of  the  square  of  a 
monomial,  we  have  the  following 

RULE. 

1.  Extract  the  square  root  of  the  coefficient. 
II.  Divide  the  exponent  of  each  letter  by  2. 

Thus,     V64a66*= 8^352     for     Sa'^b'^xSaW=zQAa%^. 

2.  Find  the  square  root  of   Q2ba%^c^.  Ans.  25a¥c^. 

3.  Find  the  square  root  of  576a*5^c^.  Ans.  24a^^c^. 

4.  Find  the  square  root  of   1960:^^2^^.  ^1??^.  l^x^yz"^, 

5.  Find  the  square  root  of  441  a^b^c^^d^^. 

Ans.  21  a-b^c^d^. 

6.  Find  the  square  root  of  7S4a^'^b^^c^hP. 

Ans.  2Sa%'c^d, 

7.  Find  the  square  root  of  81a^5^c^. 

Ans.  da'^b^c^. 


QuAsi*. — 103.  How  do  you  extract  the  square  root  of  a  monomial  1 


EXTRACTION  OF  THE  SQUARE  ROOT.       145 

104.  From  the  preceding  rule  it  follows,  that  when  a 
monomial  is  a  perfect  square,  its  numerical  coefficient  is  a 
perfect  square,  and  all  its  exponents  even  numbers.  Thus, 
25f<*52  is  a  perfect  square,  but  98a5*  is  not  a  perfect  square, 
because  98  is  not  a  perfect  square,  and  a  is  affected  with 
an  uneven  exponent. 

In  the  latter  case,  the  quantity  is  introduced  into  the  cal- 
culus by  affecting  it  with  the  sign  -y/         ,  and  it  is  written 

thus :  

^/9Sab\ 

Quantities  of  this  kind  are  called  radical  quantities,  or  irra- 
tional quantities,  or  simply  radicals  of  the  second  degree. 
They  are  also,  sometimes  called  Surds, 

These  expressions  may  often  be  simplified,  upon  the  prin- 
ciple that,  the  square  root  of  the  product  ofttoo  or  more  factors 
is  equal  to  the  product  of  the  square  roots  of  these  factors ;  or, 
in  algebraic  language, 

•yjahcd  .  .  .   =zy/ a  ,  ^h  .  -^ c  .  ^d  .  .  . 

This  being  the  case,  the  above  expression,  VoSo^^,  can 
be  put  under  the  form 

■y/\9¥x2a=:  ^/A9¥x  ^/2a, 

Now  ^/A^h^  may  be  reduced  to  7Z>2 ;  hence, 

-v/98^=:  7*2^2^ 
In  like  manner, 

^/AbaWM:=z  y/^Wc^x^bd^: Zahc ^553", 

y/B6W^c^=z  ^/XiA^b^c^^xUcz=zl2ah'^c^ ^/Wc. 
13 


146  FIRST  LESSONS  IN  ALGEBRA. 

The  quantity  which  stands  without  the  radical  sign  is 
called  the  coejfficient  of  the  radical.     Thus,  in  the  expressions 

the  quantities  75^,  ^ahc,  I2ab'^c^,  are  called  coefficients  of 
the  radicals. 

Hence,  to  simplify  a  radical  expression  of  the  second 
degree,  we  have  the  following 

RULE. 

I.  Separate  the  expression  into  two  parts,  of  which  one  shall 
contain  all  the  factors  that  are  perfect  squares^  and  the  other 
the  remaining  ones. 

II.  Take  the  roots  of  the  perfect  squares  a?id  place  them 
before  the  radical  sign,  under  which  leave  those  factors  which 
are  not  perfect  squares. 

O  105.  Remark. — To  determine  if  a  given  number  has 
any  factor  which  is  a  perfect  square,  we  examine  and  see 
if  it  is  divisible  by  either  of  the  perfect  squares 

4,     9,     16,     25,     36,     49,     64,     81,  &c ; 

and  if  it  is  not,  we  conclude  that  it  does  not  contain  a  fac- 
tor which  is  a  perfect  square. 


Quest. — 104.  When  is  a  monomial  a  perfect  square  1  When  it  is 
not  a  perfect  square,  how  is  it  introduced  into  the  calculus  1  What  are 
quantities  of  this  kind  called  1  May  they  be  simplified  \  Upon  what 
principle  '?  What  is  a  coefficient  of  a  radical  1  Give  the  rule  for  reducing 
radicals. — 105.  How  do  you  determine  whether  a  given  number  has  a 
factor  which  is  a  perfect  square  1 


EXTRACTION    OF    THE    SQUARE    ROOT.  147 

EXAMPLES. 

1.  Reduce      -y/lWiFbc     to  its  simplest  form. 

Ans.  5a^^3aEc. 

2.  Reduce      ■\/l28¥a^     to  its  simplest  form. 

Ans,  Sh'^a?d^/2f. 

3.  Reduce      -\/32«^Z>^c     to  its  simplest  form.         '   ""- 

Ans.  Aa'^¥'y/2ac, 

4.  Reduce      -y/'IsSo^p^    ^q  j^g  simplest  form. 

Ans.  IGai^c*. 

5.  Reduce      '\/\02Aa^^    to  it's  simplest  form.  w 

Ans.  22a'^b^c^ -y/Wc. 

6.  Reduce      'y/l29d^h^c^d     to  its  simplest  form. 

Ans.  2'7aWc'^^/^I, 

7.  Reduce      ^/Qlba'h^c^d     to  its  simplest  form. 

Ans.  l5a^Pc^3abd. 

8.  Reduce      ^/]A45a^c^    to  its  simplest  form. 

Ans.  17  ac^d^-x/da. 

9.  Reduce      ^/l008a^d''m^    to  its  simplest  form. 

Ans.   Ua'^d^m'^ 'x/TaJ, 
10.  Reduce      ■\/2l 56a^%^'c^    to  its  simplest  form. 

Ans.  Ua^b^c^^TT. 


11.  Reduce     y/~i05a^b^d^    to  its  simplest  form. 

Ans.  da^Pd'^^^^. 


148  FIRST  LESSONS  IN  ALGEBRA. 

106.  Since  like  signs  in  both  the  factors  give  a  plus 
sign  in  the  product,  the  square  of  — «,  as  well  as  that  of 
+  «,  will  be  «2  J  hence  the  root  of  a"  is  either  -\-a  or  —a. 
Also,  the  square  root  of  25a^^  is  either  -]-5ab'^  or  —5ab^. 
Whence  we  may  conclude,  that  if  a  monomial  is  positive, 
its  square  root  may  be  affected  either  with  the  sign  -|-  or 
—  ;  thus,  V^a^^dtSa^  ;  for,  -j-Sa^  or  —Sa^,  squared, 
ffives  9a^.  The  double  sioni  ±  with  which  the  root  is 
affected  is  read  plus  ar  minus. 

Ift-the  proposed  monomial  were  iiegative,  it  would  be  im* 
possible  to  extract  its  root,  since  it  has  just  been  shown  that 
the  square  of  every  quantity,  whether  positive  or  negative, 
is  essentially  positive.     Therefore, 

are  algebraic  symbols  which  indicate  operations  that  cannot 
be  performed.  They  are  called  imaginary  quantities^  or 
rather  imaginary  expressions,  and  are  frequently  met  with 
in  the  resolution  of  equations  of  the  second  degree.  These 
symbols  can,  however,  by  extending  the  rules,  be  simplified 
in  the  same  manner  as  those  irrational  expressions  which 
indicate  operations  that  cannot  be  performed.  Thus,  V— 9 
may  be  reduced  by  (Art.  104).     Thus, 

and  ^/'^^^Aa^=:  ^/a^ X  ^/^-i  =2a ^/^^iTi  also, 

V  — 8a2^,z=  'y/Aa^X—2:h=:2a ^/^^2b  =  2a ^/Wx  ■y/'^^^. 


Quest. — 106.  What  sign  is  placed  before  the  square  root  of  a  monor- 
mian  Why  may  you  place  the  sign  plus  or  minus  1  What  is  an  ima- 
ginary quantity  *?     Why  is  it  called  imaginary  1 


RADICALS    OF    THE    SECOND    DEGREE  149 


Of  the  Calculus  of  Radicals  of  the  Second  Degree. 

107.  A  radical  quantity  is  the  indicated  root  of  an 
imperfect  power. 

The  extraction  of  the  square  root  gives  rise  to  such  ex- 
pressions as  -y/a^  3  V^,  7  ^2,  which  are  called  irra-^ 
tional  quantities,  or  radicals  of  the  second  degree.  We  will 
now  establish  rules  for  performing  the  four  fundamental 
operations  on  these  expressions. 

108.  Two  radicals  of  the  second  degree  are  similar, 
when  the  quantities  under  the  radical  sign  are  the  same  in 
both.  Thus,  3  -y/^  and  5c  ^/b  are  similar  radicals  ;  and 
so  also  are    9  ^2    and    7  ^/2. 


Addition. 

109.    Radicals  of  the  second  degree  may  be  added 
together  by  the  following 

RULE. 

I.  If  the  radicals  are  similar  add  their  coefficients,  and  to 
the  sum  annex  the  common  radical, 

II.  If  the  radicals  are  not  similar,  connect  them  together 
with  their  proper  signs. 

Thus,  ^a^/h+^c^/h  =  {^a+bc)^/h. 


Quest. — 107.  What  is  a  radical  quantity  1  What  are  such  quantities 
called  1 — 108.  When  are  radicals  of  the  second  degree  similar  1 — 
109.  How  do  you  add  similar  radicals  of  the  second  degree  1  How  do 
you  add  radicals  which  are  not  similar  1 

13* 


150  FIRST  LESSONS  IN  ALGEBRA. 

In  like  manner. 

Two  radicals,  which  do  not  appear  to  be  similar  at  first 
sight,  may  become  so  by  simplification  (Art.  104). 
For  example, 

^4Sab^-^b  '\/75a=4b  -|/3«+56  ^/3a  —  9h  ^/3a; 

and  2^/45'+3^=6^/b  +  3^/'^=z9^/5. 

When  the  radicals  are  not  similar,  the  addition  or  sub- 
traction can  only  be  indicated.  Thus,  in  order  to  add 
3  '\/b     to     5  "v/a,     we  write 

EXAMPLES. 

1.  What  is  the  sum  of     •y/WcT'  and     V^SoM 

Ans,  Ta-y/S. 


2.  What  is  the  sum  of     ^/ma^b'^     and      y/'72a^b'^  1 

Ans.   l\a%'y/2'. 

3.  What  is  the  sum  of   W  and     \/-       ■  ? 


Ans.  4a  \, 


4.  What  is  the  sum  of     Vl25     and     V^OOa^  ? 

Ans.  (5  +  10a)V57 


RADICALS    OP    THE    SECOND    DEGREE.  151 


„r,     .    ,  ^     /  50         ,      /Too  , 

5.  What  IS  the  sum  of  \/     ,^      and    \/  ^^^    ? 


147  V   294 


6.  What  is  the  sum  of    VMS*    and      V36x2--36a2? 

j4w5.  '1a^J^Zx-\-^^JW—a^. 

7.  What  is  the  sum  of    V^^^^^     3,nd      V^SSo*^? 

^W5.  (7a+12«2a:2)V2Z 
8    Required  the  sum  of    ^^72"    and      -v/^^^- 

Ans.  14  V^ 

9.  Required  the  sum  of    -y^     and     -v^l47. 

^;i;y.   10  -y/'S"^ 

/2^  /~27" 

10.  Required  the  sum  of -i/—-    and 


3  V     50 

Ans.   ^V~6*. 

1 1 .  Required  the  sum  of  2  'y/o^J    and     3  ^64^0?^. 

Ans.  (2a+24a:2)y^T: 

12.  Required  the  sum  of    ^243     and     lO-y/sesT 

u4n5.  119V^ 

13.  What  is  the  sum  of     -y/S^Oo^P     and      VSTS^sp  ? 

^n^.  (8a5+7«^i3)y^, 

14.  What  is  the  sum  of     VtSo^F     and     -x/SOOa^^  ? 


152  FIRST  LESSONS  IN  ALGEBRA, 

Subtraction. 

1 1 0.  To  subtract  one  radical  expression  from  another, 
we  have  the  following 

RULE. 

I.  If  the  radicals  are  similar,  subtract  their  coejfficients^ 
and  to  the  difference  annex  the  common  radical. 

II.  If  the  radicals  are  not  similar,  their  difference  can  only 
he  indicated  hy  the  minus  sign, 

EXAMPLES. 

1.  What  is  the  difference  between    ^a^fb    and    a^l  ? 

Here    ^a'}/~b—a'\/~bz=z2a^\fb7Ans. 

2.  From27«V27F  subtract    Qa^/~27¥7 

First,     27av57P~=27a5VX  and    6a  V^TP^lSa^^V^ 
and  'Z^ah^J~^  —  \Qah^/~^z=z9ah^/~^    Ans. 

3.  What  is  the  difference  of    ^/~Tb    and     V^  • 

A71S.    -y/  3  . 

4.  What  is  the  difference  of    ^~2i^¥  and    V^^F"? 

Ans.   (2a5— 362)ynr 


Quest. — 110.  How  do  you  subtract  similar  radicals  1     How  do  you 
subtract  radicals  which  are  not  similar  ? 


RADICALS    OF    THE    SECOND    DEGREE.  163 


ip:  5.  Required  the  difference  of  \  / —     and    \  /— -. 

^  ^  V    5  V  27 

4      . 

Ans.  -— vl5. 
45 


6.  What  is  the  difference  of    vn[28a352    and     '}/32a^  1 

Ans.  {8ab—4:a'^)^/2a. 

7.  What  is  the  difference  of    ^/4:8a^b^    and     -x/dab    ? 

Ans.  4ab\^3ab — 3  -^  ab. 

8.  What  is  the  difference  of   'v/24~2«^^    and     '\/2a^Wl 

Ans.   {na^^-ab)^^/2^. 


>f^9.  What  is  the  difference  of  \/- 


—     and    \/—r-  ? 


Ans.  -^V~^' 


10.  What  is  the  difference  of    ^320^2    and     -x/SOa^  ? 

Ans.  4a^  5  , 

1 1 .  What  is  the  difference  between 

'v/720a3p     and      y^5ab^'^  ? 

Ans.  {I2ab—7cd)^5ab, 

12.  What  is  the  difference  between 

V968^2j2     and      '}/200a'^b^  ? 

Ans.  12ab\/2, 

13.  What  is  the  difference  between 

'v/ll2a8Z>6     and      -y/gSa^  ? 

Ans,  2a*J3'v/7r 


154  FIRST  LESSONS  IN  ALGEBRA. 


Multiplication. 

111.  For  the  multiplication  of  radicals,  we  have  the 
following 

RULE. 

I.  Multiply  the  quantities  under  the  radical  signs  together, 
and  place  the  common  radical  over  the  product. 

II.  If  the  radicals  have  coefficients,  ice  multiply  them  to- 
gether, and  place  the  product  before  the  common  radical. 

Thus,  -y/lt  X  V'^=  V^  5 

This  is  the  principle  of  Art.   104,  taken  in  the  inverse 
order. 


EXAMPLES. 

1 .  What  is  the  product  of    3  ^/bah    and    4  -y/^Oa'l 

Ans.  I20a^/T, 

2.  What  is  the  product  of   la^hc    and    '^a^/~hc'\ 

Ans.  6a%c. 

3.  What  is  the  product  of  2a  ^/c^-fW'  and   —  3a  V^^+F  ? 

Ans.   —Qa^a^+I)^), 


Quest. — 111.  How  do  you  multiply  quantities  which  are  under  radi- 
cal signs  1  "When  the  radicals  have  coefficients,  how  do  you  multiply 
them  % 


RADICALS    OF    THE    SECOND    DEGREE.  155 

4.  What  is  the  product  of    3  V"2~  and    2  -y/WJ 

A71S.  24. 

5.  What  is  the  product  of   f  ^/^a^b    and    ^^  -y/^c'^b  1 

Ans.   ^^ahc  yTS^ 

6.  What  is  the  product  of   2a?  +  -\/T    and    2x—  ^/l^  1 

Ans.  Ax^ — b. 

7.  What  is  the  product  of 


-y/a^^^/b     and      ^/ a—2^/b\ 

Ans.   -y/W—Ab. 
8.  What  is  the  product  of     '^a-^fa^     by     -y/^  ? 

Ans.  9a^y^ 

Division, 

112.  To  divide  one  radical  by  another,  we  have  the 
following 

RULE. 

I.  Divide  one  of  the  quantities  under  the  radical  sign  by  the 
other,  and  place  the  common  radical  over  the  quotient. 

II.  If  the  radicals  have  coefficients,  divide  the  coejfficieiit  of 
the  dividend  by  the  coefficient  of  the  divisor,  and  place  the 
quotient  before  the  common  radical. 


Quest. — 112.  How  do  you  divide  quantities  which  are  under  the 
radical  sign  1  When  the  radicals  have  coefficients,  how  do  you  divide 
theml 


156  FIRST    LESSONS    IN    ALGEBRA. 

Thus,     — =rzi\/-—  ;     for  the    squares  of  these  two 

expressions  are  equal  to  the  same  quantity     —  ;     hence 
the  expressions  themselves  must  be  equal. 

EXAMPLES. 

1.  Divide     ba^Jh     by     ^h-yfc.  Arts,  -^x/ — • 

2.  Divide     \2ac-\/Wc     by     4c ^2?^  Ans.  '^ayJYc. 

3.  Divide     G^V^GF    by     SV^F.  Ans.  \ahyj^. 

4.  Divide     4a2y^5oF    by     2a^  ^/W,        Jlus.  2Z>2yTo; 

5.  Divide     "Z^a^hyJ^^^     by     13a  ^9^?^ 

Ans.  Qa^b  y/ah. 

6.  Divide     %\a%^yJWac     by     \1ab  yj^a, 

Ans.   ^d^y^  ^/c, 

7.  Divide      y/^cP-     by      -ypZ.  Ans.  \a. 

8.  Divide     ^aWypZ^'^^     by     12^5^.  Ans.  o?}P: 

9.  Divide     ^ayflW  by     3'v/"57  .4^^.  2ab^/2: 

10.  Divide     iSb^^W   by     2b'^^^'.  Ans.  360b^. 

11.  Divide     Sa^^^c^  ^73^     ^y     2a'x/2M: 

Ans.  2a¥c^d, 

12.  Divide     ^Qa^c^y/dW    by     48aZ»cV'2?T 


RADICALS    OF    THE    SECOND    DEGREE.  157 

13.  Divide     21  a^h^ -y/^lU?     by      ^J'U. 

Arts,  27^656 -y/3. 

14.  Divide     ISaW^gJ?     by     ^ah^fcF: 

Ans.  ^a%^  ^J2. 

To  Extract  the  Square  Root  of  a  Polynomial, 

113.  Before  explaining  the  rule  for  the  extraction  of 
the  square  root  of  a  polynomial,  let  us  first  examine  the 
squares  of  several  polynomials  :  we  have 

{a+hf  —  a?+2al-\-h'^, 

{a+h+cY=a?+2ah+P-^2{a+h)c+c\ 

{a+h-\-c-\-df—a?+2ah  +  h'^+2{a+h)c+c^ 

^2{a^h  +  c)d+d?. 

The  law  by  which  these  squares  are  formed  can  be  enun- 
ciated thus  : 

The  square  of  any  polynomial  contains  the  square  of  the 
first  term,  plus  twice  the  product  of  the  first  term  hy  the  second^ 
plus  the  square  of  the  second ;  plus  twice  the  first  two  terms 
multiplied  hy  the  third,  plus  the  square  of  the  third  ;  plus  twice 
the  first  three  terms  multiplied  hy  the  fourth,  plus  the  square 
of  the  fourth ;  and  so  on» 


Quest. — 113.  What  is  the  square  of  a  binomial  equal  to]  What 
is  the  square  of  a  trinomial  equal  to  1  What  is  the  square  of  any 
polynomial  equal  to  1 

14 


158  FIRST  LESSONS  IN  ALGEBRA. 

114.  Hence,  to  extract  the  square  root  of  a  polynomial 
we  have  the  following 

RULE. 

I.  Arrange  the 'polynomial  with  reference  to  one  of  its  letters 
and  extract  the  square  root  of  the  first  term  :  this  will  give  the 
first  term  of  the  root. 

II.  Divide  the  second  term  of  the  polynomial  hy  double  the 
first  term  of  the  root,  and  the  quotient  will  be  the  second  term 
of  the  root. 

III.  Then  form  the  square  of  the  two  terms  of  the  root 
found,  and  subtract  it  from  the  first  polynomial,  and  then 
divide  the  first  term  of  the  remainder  by  double  the  first  term 
of  the  root,  and  the  quotient  will  be  the  third  term. 

IV.  Form  the  double  products  of  the  first  and  second  terms, 
by  the  third,  plus  the  square  of  the  third ;  then  subtract  all 
these  products  from  the  last  remainder,  and  divide  the  first 
term  of  the  result  by  double  the  first  term  of  the  root,  and  the 
quotient  will  be  the  fourth  term.  Then  proceed  in  the  same 
manner  to  find  the  other  terms. 

EXAMPLES. 

1 .  Extract  the  square  root  of  the  polynomial 
49«2^2_24aZ/3  4.25a4_30«35-]_1654. 
First  arra,nge  it  with  reference  to  the  letter  a. 


25«^  —  30^3^ + A9d^b'^—2Aab^  -f  1 65-^ 
25a4_30a3Z>+   9a%'^ 


10^2 


40a2^2_24a53_|_i6Z,4     1st  Rem. 
40a252_24a53_|_i6Z>^ 

0     .     .     .      2d  Rem, 


^  RADICALS    OF    THE    SECOND    DEGREE  159 

After  having  arranged  the  polynomial  with  reference  to  a, 
extract  the  square  root  of  25a*,  this  gives  Sa^,  which  is 
placed  at  the  right  of  the  polynomial ;  then  divide  the 
second  term,  —  30a^Z>,  by  the  double  of  ba^,  or  lOa^  ;  the 
quotient  is  — 3ah,  and  is  placed  at  the  right  of  ba?.  Hence, 
the  first  two  terms  of  the  root  are  baF'—'iah.  Squaring  this 
binomial,  it  becomes  25a*— SOa^^+^a^^^,  which,  subtracted 
from  the  proposed  polynomial,  gives  a  remainder,  of  which 
the  first  term  is  4:00?})^.  Dividing  this  first  term  by  \Oa?j 
(the  double  of  ba?'),  the  quotient  is  +Ah'^  ;  this  is  the  third 
term  of  the  root,  and  is  written  on  the  right  of  the  first  two 
terms.  By  forming  the  double  product  of  ba?  —  3ah  by  4:h'^, 
and  at  the  same  time  squaring  4Z>2,  we  find  the  polynomial 
A0a^h'^—2Aah^-\-\Qh^,  which,  subtracted  from  the  first  re- 
mainder, gives  0.  Therefore  5a?  —  ^ah-\-Ah'^  is  the  required 
root. 

2.  Find  the  square  root  of      a'^-{-Aa'^x+Qa"x^-\-Aax'^-\-x^, 

Ans.  aF'-{-2ax-{-x'^. 

3.  Find  the  square  root  of      a^—Aa^x+Qa?x'^~Aax'^+x^. 

Ans,  aF- — 2ax-\-x'^, 

4.  Find  the  square  root  of 

Ax^-\-\2x^+bx^—2x'^+lfx'^—2x+l. 

Ans,  2x^+^x'^—x+l. 

5.  Find  the  square  root  of  * 

9a*-12a35+28«2^2_i6«^>3+16K 

Ans,  3a^—2ab+4b^. 


Quest. — 114.  Give  the  rule  for  extractxig  the  square  root  of  a  poly- 
nomial 1  What  is  the  first  step  'l  What  the  second  1  What  the  third  '^ 
What  the  fourths 


160  FIRST  LESSONS  IN  ALGEBRA. 

6.  What  is  the  square  toot  of 

Ans.  a?2— 2«a?— 2. 

7.  What  is  the  square  root  of 

9^2 — 12a;+6a?y  +  y^ — 4y  +  4. 

Ans.  Sic+y— 2. 

8.  What  is  the  square  root  of    y^—2y'^x'^-^2x'^—2i/-{-l 

'  ^  •  Ans.  y'^—x'^  —  l, 

9.  What  is  the  square  root  of    9a^6^—30«3Z>3+ 25^2^2  ? 

Ans.  Sa^b^—dab, 

10.  Find  the  square  root  of 
2ba^b^—40a^^c-{-76a'^b^c^—48ab^c^-{-36b^c'^—30a^bc 

+ 24a^bc^ — 36a2&c3 + ^a^c^. 

Ans.  ba?b  —  3<i^c — 4abc  -\-Qbc^. 

115.  We  will  conclude  this  subject  with  the  following 
remarks. 

1st.  A  binomial  can  never  be  a  perfect  square,  since  we 
know  that  the  square  of  the  most  simple  polynomial,  viz : 
a  binomial,  contains  three  distinct  parts,  which  cannot  ex- 
perience any  reduction  amongst  themselves.  Thus,  the 
expression  a?-\-b'^  is  not  a  perfect  square  ;  it  wants  the  term 
±2ab  in  order  that  it  should  be  the  square  oi  a±b. 

2nd.  In  order  that  a  trinomial,  when  arranged,  may  be  a 
perfect  square,  its  two  extreme  terms  must  be  squares,  and 
the  middle  term  must  be  the  double  product  of  the  square 
roots  of  the  two  others.  Therefore,  to  obtain  the  square 
root  of  a  trinomial  when  it  is  a  perfect  square  ;  Extract  the 
roots  of  the  two  extreme  terms,  and  give  these  roots  the  same 
or  contrary  signs,  according  as  the  middle  term  is  positive  or 


RADICALS    OF    THE    SECOND    DEGREE.  161 

negative.      To  verify  it,  see  if  the  double  product  of  the  two 
roots  gives  the  middle  term  of  the  trinomial.     Thus, 

9a^—48a^Z>2-j- 64^254     is  a,  perfect  square, 

since  -^/d^^Sa^,  and    ^/MM^=—8ab\ 

and  also     2  X  3a^  X  — 8ab^=  —4Sa^b^=  the  middle  term. 

But  4a^-\-l4ab-\-9b^  is  not  a  perfect  square  :  for  although 
4a"  and  4-9b^  are  the  squares  of  2a  and  3b,  yet  2  X2ax  36 
is  not  equal  to  I4ab. 

3rd.  In  the  series  of  operations  required  in  a  general  ex- 
ample, when  the  first  term  of  one  of  the  remainders  is  not 
exactly  divisible  by  twice  the  first  term  of  the  root,  we  may 
conclude  that  the  proposed  polynomial  is  not  a  perfect 
square.  This  is  an  evident  consequence  of  the  course  of 
reasoning,  by  which  we  have  arrived  at  the  general  rule  for 
extracting  the  square  root. 

4th.  When  the  polynomial  is  not  a  perfect  square,  it  may 
be  simplified  (See  Art.  104.) 

Take,  for  example,  the  expression     '\/a^-\-4a'^b^-{-4ab^. 

The  quantity  under  the  radical  is  not  a  perfect  square  ; 
but  it  can  be  put  under  the  form  ab{a'^-\-4ab-\-4b'^).  Now, 
the  factor  between  the  parenthesis  is  evidently  the  square 
of  a -\- 2b,  whence  we  may  conclude  that, 

/'  ^a^  +  4a^b'^+4ab^  =  {a-\-2b)  VoF. 

2.  Reduce"  2a2Z>— 4a J2_j_ 253    to  its  simple  form. 

Ans.  {a  —  b)^2F. 

Quest. — 115.  Can  a  binomial  ever  be  a  perfect  power'?  "WTiy  not  1 
When  is  a  trinomial  a  perfect  power  1  When,  in  extracting  the  square 
root  we  find  that  the  first  term  of  the  remainder  is  divisible  by  twice  the 
root,  is  the  polynomial  a  perfect  power  or  not  1 

14* 


162  FIRST  LESSONS  IN  ALGEBRA. 


CHAPTER  VL 

Equations  of  the  Second  Degree, 

116.  An  equation  of  the  second  degree  is  an  equation 
involving  the  second  power  of  the  unknown  quantity,  or  the 
product  of  two  unknown  quantities.     Thus, 

aj^zzrcr,     ax^-\-hxzz::c,     and     xyz^zd?, 

are  equations  of  the  second  degree. 

117.  Equations  of  the  second  degree  are  of  two  kinds, 
viz :  equations  involving  two  terms,  which  are  called  incom- 
plete equations  ;  and  equations  involving  three  terms,  which 
are  called  complete  equations.     Thus, 

x^=:a     and     ax^  =:  h, 

are  incomplete  equations  ;  and 

x^-\-2axz=zb,     and     ax'^-\-bx=zdf 

are  complete  equations. 


Quest. — 116.  What  is  an  equation  of  the  second  degree  1 — 117.  How 
many  kinds  aie  there  1  What  is  an  incomplete  equation  1  What  is  a 
complete  equation  1 


EQUATIONS  OF  THE  SECOND  DEGREE      163 

118.  When  we  speak  of  an  equation  involving  tvv^o 
terms,  and  of  an  equation  involving  three  terms,  we  under- 
stand that  the  equation  has  been  reduced  to  its  simplest 
form. 

Thus,  if  we  have  the  equation 

although  in  its  present  form  there  are  four  terms,  yet  it  may- 
be reduced  to  an  equation  containing  but  two.  For,  by- 
adding  3x^  to  4a;2  and  transposing  —4,  we  have 

7ic2:=10. 

Also,  if  we  have 

3x'^-i-5x-{-7x+5  —  9, 

we  get  by  reducing 

3x'^-^l2x=z4, 

an  equation  containing  but  three  terms. 
Again,  if  we  take  the  equation 

ax^+bx^+d=:f 


we  have 


(a-{-b)x^=f-d     and     a;2r=£-^, 


an  equation  of  two  terms. 


Quest. — 118.  When  you  speak  of  an  equation  involving  two  terms, 
do  you  speak  of  the  equation  after  it  has  been  reduced,  or  before  1  When 
you  speak  of  an  equation  of  three  terms,  is  it  the  reduced  equation  to 
which  you  refer '?  To  what  forms,  then,  may  every  equation  of  the  second 
degree  be  reduced  1 


164  FIRST    LESSONS    IN    ALGEBRA. 

Also,  if  we  have       ax^ + dx""  ■\-fx + 5 = c 
we  obtain  {a-\-d)x^-\-fx=.c—'b, 

and  consequently 


a-\-d       a-\-d^ 

an  equation  of  three  terms. 

Hence  we  may  conclude  :  That  every  equation  of  the 
second  degree  may  he  reduced  to  an  incomplete  equation  involv' 
ing  two  terms,  or  to  a  complete  equation  involving  three  terms. 

Of  Inco7nplete  Equations, 

1 .  What  number  is  that  which  being  multiplied  by  itself 
the  product  will  be  144. 

Let  X—  the  number :  then 

a:x  xzzzx'^zizlii. 

It  is  plain  that  the  value  of  x  will  be  found  by  extracting 
the  square  root  of  both  members  of  the  equation  :  that  is 

'^x'^zzi  -1/144  :  that  is,  a:  =12. 

2.  A  person  being  asked  how  much  money  he  had,  said 
if  the  number  of  dollars  be  squared  and  6  be  added,  the  sum 
will  be  42  :  How  much  had  he  ? 

Jjct  x=z  the  number  of  dollars. 

Then  by  the  conditions 

hence,  x'^=z42  —  6  =  36     and     xz=e. 

Ans.  $6. 


EQUATIONS    OF    THE    SECOND    DEGREE.  166 

3.  A  person  being  asked  his  age  said,  if  from  the  square 
of  my  age  you  take  192,  the  remainder  will  be  the  square 
of  half  my  age  :  what  was  his  age  1 

Denote  his  age  by  x. 

Then,  by  the  conditions  of  the  question 


-192: 


<y'>4. 


and  by  clearing  the  fractions 

4x^—768=x^; 
hence,  4a;2— a;2=:768, 

and  3x^=768 

0:2=256 
X  =    16. 


Ans.  16. 


119.  There  is  no  difficulty  in  the  resolution  of  an  equa- 
tion  of  the  form    ax'^=b.     We    deduce   from   it    x^= — , 

a 


whence 


^=Vt- 


When    —    is  a  particular  number,  either  entire  or  frac- 
a 


tional,  we  can  obtain  the  square  root  of  it  exactly,  or  by  ap- 
proximation. If  —  is  algebraic,  we  apply  the  rules  estab- 
lished for  algebraic  quantities. 

QuBst. — 119.  How  do  you  resolve  an  incomplete  equation  1 


166  FIRST  LESSONS  IN  ALGEBRA. 

Hence,  to  find  the  value  of  x  we  have  the  following 

RULE. 

I.  Find  the  value  of'^. 

II.  Then  extract  the  square  root  of  both  members  of  the 
equation. 

4.  What  is  the  value  of  x  in  the  equation 
3a:H-8=:5a?2— 10. 

By  transposition     3a;2 — 5a?2  —  —  1 0  —  8, 
by  reducing  —2x'^z=:i  —  \B, 

by  dividing  by  2  and  changing  the  signs 

by  extracting  the  square  root     a;=:3. 

We  should,  however,  remark  that  the  square  root  of  9, 
is  either     +3,     or     —3.     For, 

+  3x+3  =  9     and     — 3x— 3=9. 

Hence,  when  we  have  the  equation 

we  have  £c=:  +  3     and     a?=:~3. 

1 20.  A  root  of  an  equation  is  such  a  number  as  bemg 
substituted  for  the  unknown  quantity,  will  satisfy  the  equa- 
tion, that  is,  render  the  two  members  equal  to  each  other. 
Thus,  in  the  equation 


there  are  two  roots,    +3    and    —  3  ;     for  either  of  these 
numbers  being  substituted  for  x  will  satisfy  the  equation. 


EQUATIONS    OF    THE    SECOND    DEGREE. 

5.  Again,  if  we  take  the  equation 

we  shall  have 

a;=  +  \/ —     and     x^=.  —  \/ — . 
^    a  y     a 


167 


For, 


=:h,     or     ax  —  =  b, 
a 


and 


ax 


=  £>,     or     aX — =6. 
a 


Hence  we  may  conclude, 

1st.   That  every  incomplete  equation  of  the  second  degree 
has  two  roots. 

2nd.   That  those  roots  are  numerically  equal  but  have  con- 
trary signs. 

6.  What  are  the  roots  of  the  equation 

3a:2+6  — 4a:2_10. 

Ans.  x=:-\-4    and   x=z—4, 

7.  What  are  the  roots  of  the  equation 

1  x^ 

0:2-8=::  — +10. 

3  9 

Ans.  a?™ -{-9    and   x^z—d. 


Quest. — 120.  What  is  the  root  of  an  equation^  What  are  the  roots 
of  the  equation  a:2  =  9  1  Of  the  equation  ax2=b  1  How  many  roots  has 
every  incomplete  eqvation  1  How  do  those  roots  compare  with  each 
other  1 


168  FIRST  LESSONS  IN  ALGEBRA. 

8.  What  are  the  roots  of  the  equation 

6ic2— 7  =  3a;2+5. 

Ans.  x=+2,     a:=~2. 

9.  What  are  the  roots  of  the  equation. 

8+5x^  =  ^  +  4x^+28, 

0 

Ans.  x=:+5j     x='—5, 

10.  Find  a  number  such  that  one-third  of  it  multiplied 
by  one-fourth  shall  be  equal  to  108  ? 

Ans.  36. 

1 1 .  What  number  is  that  whose  sixth  part  multiplied  by 
its  fifth  part  and  product  divided  by  ten,  shall  give  a  quo- 
tient equal  to  3  ? 

Ans.  30. 

12.  What  number  is  that  w^hose  square,  plus  18,  shall  be 
equal  to  half  its  square  plus  30^. 

Ans.  5. 

13.  What  numbers  are  those  v^hich  are  to  each  other  as 
1  to  2  and  the  difference  of  v^hose  squares  is  equal  to  75. 

Let        x=:     the  less  number. 

Then   2x=z     the  greater. 

Then  by  the  conditions  of  the  question 

4a;2— a;2=75, 

hence,  3x'^=75  ; 

and  by  dividing  by     3,    x'^=:25     and     x=5, 

and  2a?=10. 

Ans.  5  and  10. 


EQUATIONS    OF    THE    SECOND   DEGREE.  169 

;•- 

14.  What  two  numbers  are  those  which  are  to  each  other 
as  5  to  6,  and  the  difference  of  whose  squares  is  44. 

Let  x=:    the  greatest  number. 

5 

Then  -^x=    the  least. 
6 

By  the  conditions  of  the  question 
.25 


x^- 

~— -a?  =44. 
36 

by  clearing 

fractions, 

36a?2- 

-25a?2=1584; 

hence, 

lla;2=1584. 

and 

a?2=144. 

hence, 

X  =12, 

and 

4.  =10. 

Ans.  10  and  12, 

15.  What  two  numbers  are  those  which  are  to  each 
other  as  3  to  4,  and  the  difference  of  whose  squares  is  28  ? 

Ans.  6  and  8. 

16.  What  two  numbers  are  those  which  are  to  each  other 
as  5  to  1 1 ,  and  the  sum  of  whose  square  is  584  ? 

Ans.   10  and  22. 

17.  A  says  to  B,  my  son's  age  is  one  quarter  of  yours, 
and  the  difference  between  the  squares  of  the  numbers  re- 
presenting their  ages  is  240 :  what  were  their  ages  ? 

.        (Eldest      16. 
Ans.  <  ^^ 

(  Younger    4. 

15 


170  FIRST  LESSONS  IN  ALGEBRA. 

When  there  are  two  unknown  quantities, 

121.  When  there  are  two  or  more  unknown  quantities, 
eliminate  one  of  them  by  the  rule  of  Article  77  :  there  will 
thus  arise  a  new  equation  with  but  a  single  unknown  quantity, 
the  value  of  which  may  be  found  by  the  rule  already  given. 

1.  There  is  a  room  of  such  dimensions,  that  the  differ- 
rence  of  the  sides  multiplied  by  the  less  is  equal  to  36,  and 
the  product  of  the  sides  is  equal  to  360  :  what  are  the 
sides  1 

Let   x=z    the  less  side  ; 
y=z    the  greater. 

Then,  by  the  1st  condition, 

(y— a?)a;  =  36  ; 

and  by  the  2nd,  '  0:3/ =360. 

From  the  first  equation,  we  have 

icy— a;2=36  ; 

and  by  subtraction,  a:2=r324. 

Hence,  cc=^d2i'—lS\ 


'''  .20. 


18 


Ans.  a?=rl8,y=20. 


Quest. — 121.  How  do  you  resoh-e  the  equation  when  there  are  two 
or  more  unknown  quantities  1 


EQUATIONS    OF    THE    SECOND    DEGREE.  171 

2.  A  merchant  sells  two  pieces  of  muslin,  which  together 
measure  12  yards.  He  received  for  each  piece  just  so 
many  dollars  per  yard  as  the  piece  contained  yards.  Now, 
he  gets  four  times  as  much  for  one  piece  as  for  the  other  : 
how  many  yards  in  each  piece  1 

Let   x=z    the  number  in  the  larger  piece  ; 
y=    the  number  in  the  shorter  piece. 

Then,  by  the  conditions  of  the  question,        'fiitetf 

xXx=zx'^=  what  he  got  for  the  larger  piece  ; 

yXy=y'^=  what  he  got  for  the  shorter. 

And  x^zzz^y^,  by  the  2nd  condition. 

X  =^y,  by  extracting  the  square  root 

Substituting  this  value  of  x   in  the  first  equation,  we  have 

iy+y-:13; 

and  consequently,  y=   8, 

and  a?=  4. 

Ans.  8  and  4. 

3.  What  two  numbers  are  those  whose  product  is  30,  and 
quotient  3 J  ?  Ans.  10  and  3. 

4.  The  product  of  two  numbers  is  a,  and  their  quotient 
b  :  what  are  the  numbers  ? 

A71S.   -yj  ah  and  \  /  -7- . 
V     0 

5.  The  sum  of  the  squares  of  two  numbers  is  117,  and 
the  difference  of  their  squares  45  :   what  are  the  numbers  ? 

Alls,  9  and  6. 


172  FIRST    LESSONS    IN   ALGEBRA. 

6.  The  sum  of  the  squares  of  two  numbers  is  «,  and  the 
difference  of  their  squares  is  h  :  what  are  the  numbers  ? 

Ans.  x=  y^a-i-b,  y=  -y/a — b. 

7.  What  two  numbers  are  those  which  are  to  each  other 
as  3  to  4,  and  the  sum  of  whose  squares  is  225  ? 

Ans.  9  and  12. 

8.  What  two  numbers  are  those  which  are  to  each  other 

as  m  to  n,  and  the  sum  of  whose  squares  is  equal  to  a^  ? 

ma  na 

Ans.        , —     — ,        , =r. 

V?^^+^^       ym^-^-n^ 

9.  What  two  numbers  are  those  which  are  to  each  other 
as  1  to  2,  and  the  difference  of  whose  squares  is  75  ? 

Ans.  5  and  10. 

10.  What  two  numbers  are  those  which  are  to  each  other 
as  m  to  n,  and  the  difference  of  whose  squares  is  equal 
to  ^2? 

mb  nb 


Ans. 


yrn^ — n"^  '     yw 


'  m'' — n^ 

11.  A  certain  sum  of  money  is  placed  at  interest  for  six 
months,  at  8  per  cent,  per  annum.  Now,  if  the  amount  be 
multiplied  by  the  number  expressing  the  interest,  the  pro- 
duct will  be  562500  :  what  is  the  amount  at  interest  ? 

Ans.  $3750. 

12.  A  person  distributes  a  sum  of  money  between  a  num- 
ber of  women  and  boys.  The  number  of  women  is  to  the 
number  of  boys  as  3  to  4.  Now,  the  boys  receive  one- 
half  as  many  dollars  as  there  are  persons,  and  the  women 
twice  as  many  dollars  as  there  are  boys,  and  together  they 
receive  138  dollars  :  how  many  women  were  there,  and 
how  many  boys  ? 

(  36  women. 
i  48  boys. 


EQUATIONS  OF  THE  SECOND  DEGREE.      173 

Of  co7nplete  Equations. 

1 22.  We  have  already  seen  (Art.  117),  that  a  complete 
equation  of  the  second  degree,  after  it  has  been  reduced, 
contains  three  terms,  viz  :  the  square  of  the  unknown 
quantity  in  the  first  term  ;  the  first  power  of  the  unknown 
quantity  in  the  second  term  ;  and  a  known  quantity,  in  a 
third  term. 

Thus,  if  we  have  the  equation 

5a?2— 2a;24-8  =  9a;4-32, 
we  have,  by  transposing  and  reducing, 

3x'^—9x—24, 
and  by  dividing  by  3, 

x'^—3x=S, 
which  contains  but  three  terms. 
2.  If  we  have  the  equation 

a^x"^  -f  3abx  -\-x'^:=:zcx+d, 
by  collecting  the  coefficients  of  x"^  and  x,  we  have 
(^2+  l)x^+{3ab  —  c)x  =  d  ; 

and  dividing  by  the  coefficient  of  x^,  we  have 
3ab—c  d 

X^-\ —-—X=:- 


a?-\-l         d^+l 


Quest. — 122.  How  many  terms  does  a  complete  equation  of  the 
second  degree  contain  1  Of  what  is  the  first  term  composed  1  The 
second  1     The  thirds 

15* 


174  FIRST  LESSONS  IN  ALGEBRA. 

If  we  represent  the  coefficient  of  x  by  2p,  and  the  known 
term  by  q,  we  have 

an  equation  containing  but  three  terms  ;  and  we  see,  from 
the  above  examples,  that  every  complete  equation  of  the 
second  degree  may  be  reduced  to  this  form. 

1!33.  We  wish  now  to  show  that  there  are  four  forms 
under  which  this  equation  will  be  expressed,  each  depend- 
ing on  the  signs  of  2p  and  q. 

1st.  Let  us  for  the  sake  of  illustration,  make 

2p=r+4,     and     ^=+5: 

We  shall  then  have         x'^-\-4x=5. 

2nd.  Let  us  now  suppose 

2j9  =  — 4,     and     q=z-{-5  : 

we  shall  then  have         x^--4x=5. 

3rd.  If  we  make 

25'= +4,     and     $'=—5, 

we  have  x^+4xz=: — 5. 

4th.  If  we  make 

2pz=z—4,     and     q= — 5, 

we  have  x'^—bx=z—5. 


Quest. — 123.  Under  how  many  forms  may  every  equation  of  the 
second  degree  be  expressed  1  On  what  will  these  forms  depend  1  What 
are  the  signs  of  the  coefficient  of  z  and  the  known  term,  in  the  first 
form  1  What  in  the  second  1  What  in  the  third  1  What  in  the  fourth  I 
Repeat  the  four  forms. 


EQUATIONS    OF   THE    SECOND   DEGREE.  175 

We  therefore  conclude  that  every  complete  equation  of 
the  second  degree  may  be  reduced  to  one  of  these  forms  : 

a?2 + 2^a?  =^+q,  1st  form. 

a;2 — 2px = + ^,  2nd  form. 

x^ + 2px  ——q^  3rd  form. 

x^ — 2px  =  —  5',  4th  form. 

124.  Remark. — If,  in  reducing  an  equation  to  either  of 
these  forms,  the  second  power  of  the  unknown  quantity 
should  have  a  negative  sign,  it  must  be  rendered  positive 
by  changing  the  sign  of  every  term  of  the  equation. 

125.  We  are  next  to  show  the  manner  in  which  the 
*-alue  of  the  unknown  quantity  may  be  found.  We  have 
^en  (Art.  38),  that 

{x-\-pY=:x'^+2px+p'^  ; 

and  comparing  this  square  with  the  first  and  third  forms,  we 
see  that  the  first  member  in  each  contains  two  terms  of  the 
square  of  a  binomial,  viz  :  the  square  of  the  first  term  plus 
twice  the  product  of  the  2nd  term  by  the  first.  If,  then,  we 
take  half  the  coefiicient  of  x,  viz  :  p,  and  square  it,  and  add 
to  both  members,  the  equations  take  the  form 

x'^+2px-{-p'^=q  -\-p\ 

x^+2px  -}-p^  zn-^q'^  +p^i 

in  which  the  first  members  are  perfect  squares.     This  is 


Quest. — 124.  If  in  reducing  an  equation  to  either  of  these  forms  the 
coefficient  of  x^  is  negative,  what  do  you  do  ?— 125.  What  is  the  square 
of  a  binomial  equal  to  1  What  does  the  first  member  in  each  form  con- 
tain 1  How  do  you  render  the  first  member  a  perfect  square  1  What  is 
this  called  1 


176  FIRST  LESSONS  IN  ALGEBRA. 

called  completing  the  square.  Then,  by  extracting  the 
square  root  of  both  members  of  the  equation,  we  have 

and  X  -\-p  =  ±  'v/— 5'  +i^^> 

which  gives,  by  transposing  p, 

x=—p±'\/q-{-p'^, 

x=  — p  i  -y/ — q-\-p'^. 

126.  If  we  compare  the  second  and  fourth  forms  with 
the  square 

{x  — pY  =:  a;2 — 2px  +/>^, 

we  also  see  that  half  the  coefficient  of  x  being  squared  and 
added  to  both  members,  will  make  the  first  members  perfect 
squares.     Having  made  the  additions,  we  have 

ic^ — 2px  -j-p"^ = q  -\-p'^, 

x"^ — 2px  -\-p'^  =~q  +i?^- 

Then,  by  extracting  the  square  root  of  both  members,  we 
have 

and  x—pz=z  ±  ^—q+p^; 

and  by  transposing  —p^  we  find 

x=pdz^/^'p, 
and  X  =:p  dz  'y/—q-{-p'i. 


Quest. — 126.  In  the  second  form,  how  do  you  make  the  first  mem- 
ber a  perfect  square  1 


EQUATIONS    OF   THE    SECOND   DEGRlEE.  17T 

127,  Hence,  for  the  resolution  of  every  equation  of  the 
econd  degree,  we  have  the  following 

RULE, 

I.  Reduce  the  equation  to  one  of  the  known  forms. 

II.  Take  half  the  coefficient  of  the  second  term,  square  it^ 
4€nd  add  the  result  to  both  members  of  the  equation. 

III.  Then  extract  the  square  root  of  both  members  of  the 
equation ;  cfter  which,  transpose  the  known  term  to  the  second 
member. 

Remark, — The  square  root  of  the  first  member  is  always 
equal  to  the  square  root  of  the  first  term,  plus  or  minus  half 
the  coefHcient  of  x, 

EXAMPLES  IN  THE  FIRST  FORM, 

1 .  What  are  the  values  of  x  in  the  equation 

2a:24-8ir— 64  ? 
If  we  first  divide  by  the  coefficient  2,  we  obtain 
a:2+4a:=32. 
Then,  completing  the  square, 

a;2+4^-f4=32  +  4=36. 
Extracting  the  root, 

x+2  =  ±:^/ZQz=z  +  Q  or  —6. 
Hence,  a;=— 2  +  6=r-f  4  ; 

or,  x=—2~Qz=z—S. 


Quest. — 127.  Give  the  general  rule  for  resolving  an  equation  of  the 
second  degree.  What  is  the  first  step  1  What  the  second  \  What  the 
third  1     What  is  the  square  root  of  the  first  member  always  eqnal  to  1 


178  FIRST  LESSONS  IN  ALGEBRA. 

That  is,  in  this  form  the  smaller  root  is  positive,  and  the 
larger  negative. 

Verification. 

If  we  take  the  positive  value,  viz  :    a;=  +  4, 

the  equation  a:2+4a;=32 

gives  42 -{-4  X  4=:32  : 

and  if  we  take  the  negative  value  of  a?,  viz  :    xz=.  —  8, 

the  equation  x^+Ax=z^2 

gives  (-8)2+4(-8)=:64-32=:32. 

From  which  we  see  that  either  of  the  values  of  x,  viz : 
a:=+4    or    x=: — 8,    will  satisfy  the  equation. 

2.  What  are  the  values  of  x  in  the  equation 
3a;2+I2aj  — 19=— aj2_x2a;+89  ? 
By  transposing  the  terms,  we  have 

3a;2+a;2+12a:  +  12a;  =  89  +  19 ; 

and  by  reducing, 

4a;2+ 2407  =108; 

and  dividing  by  the  coefficient  of  a?^, 
a;2^6a;=27. 
Now,  by  completing  the  square, 

a;2+6a;2+9  =  36; 
extracting  the  square  root, 

a;+3  =  ±  V36'=  +  6  or  —6: 
hence,  a;=  +  6— 3  =  +  3; 

or,  «=— 6—3=— 9. 


EQUATIONS  OF  THE  SECOND  DEGREE.      179 

Verification. 
If  we  take  the  plus  root,  the  equation 

gives  (3)2  4-6(3)=27; 

and  for  the  negative  root, 

a?2+6a?=27 
gives  (-~9)2+6(~9)=:81 -54=27. 

4.  What  are  the  values  of  x  in  the  equation 

a?2— 10a?  4-15=^ 34a;+155. 

5 

By  clearing  the  fractions,  we  have 

5a;2— 50a;+75  =  a?2— 170a;+775  : 

by  transposing  and  reducing,  we  obtain 

4a;2  4-120a?=700; 

then,  dividing  by  the  coefficient  of  a;^,  we  have 

a;2+30a?=175; 

and  by  completing  the  square, 

^2+30^+225=400; 

and  by  extracting  the  square  root, 

a?+15  =  ih  V400  =  +20  or  —20. 
Hence,  a:=  +  5    or    —35. 

Verification, 
For  the  plus  value  of  x,  the  equation 
a:2+30a;=175 
gives  (5)2+30x5=25+150  =  175. 


180  FIRST  LESSONS  IN  ALGEBRA. 

And  for  the  negative  value  of  x,  we  have 

(-.35)2+30(-35)  =  1225-1050=175. 

5.  What  are  the  values  of  x   in  the  equation 

_^2  ^  _,^_.=s--x-x^+-j^  ? 

Clearing  the.  fractions,  we  have 

10a:2-6a:+9=96-8a;-12x2+273  ; 

transposing  and  reducing, 

22a;2+2ir=360; 

dividing  both  members  by  22, 

2         360 
^22  22 

/  1  \2 
Add    (  —  ]     to  both  members,  and  the  equation  becomes 

,2     .  /  1  V      360      /  1  \2 
^  +22^+122)  =-22-+V22)   ' 

whence,  by  extracting  the  square  root, 

.  1         /360  ,  /  r^ 

^+22  =  -V-22-+V22> 


therefore, 


1    ,       /360--  /  1  x2 


1\^ 


«,d  ,=  _^^_y/^^Q' 


EQUATIONS  OP  THE  SECOND  DEGREE.      181 

It  remains  to  perform  the  numerical  opwations.  In  the 
first  place,  •^+(^)  ^^^st  be  reduced  to  a  single  num- 
ber, having  (22 )'^  for  its  denominator. 

N  —     (  ^  Y_360x224-l_7921 

^^'  22"''\22/""       (22)2        -(22)2' 

extracting  the  square  root  of  7921,  we  find  it  to  be  89; 
therefore, 


'  V    22  "^V22/  - 


89 
^22- 


Consequently,  the  plus  value  of  x  is 


'^""~22'^22~22"~   ' 

and  the  negative  value  is 

___1 89____45 

^~     22     22""  ""IT' 

that  is,  one  of  the  two  values  of  x  which  will  satisfy  the 
proposed  equation  is  a  positive  whole  number,  and  the  other 
a  negative  fraction. 

6.  What  are  the  values  of  x  in  the  equation 

3a?2+2a?— 9=76. 

7.  What  are  the  values  of  x  in  the  equation 

2ir2-f8a;+7=^~^  +  197. 
4        o 


16 


182  FIRST  LESSONS  IN  ALGEBRA. 

8.  What  are  the  values  of  x  in  the  equation 


Ans.  i 


x=9 

-64^. 
9.  What  are  the  values  of  x  in  the  equation 

~-— -8-— -7^4-61 
14^-2      ^''+^2- 

c  x=z2 

(00= -^7 1 

10.  What  are  the  values  of  x  in  the  equation 


Ans, 


x^       X      x^       a?      1 3 


Ans,   \ 

i  x=.-- 


X=:\ 

2*. 


EXAMPLES    IN    THE    SECOND    FORM. 


1.  What  are  the  values  of  x  in  the  equation 
a?2  — 8a;+10=:19. 
By  transposing, 

a;2-8a:=19-10=9, 
then  by  completing  the  square 

a;2>-.8ir+ 16  =  9  +  16=25, 
and  by  extracting  the  root 

a;— 4=zfc  V25'=  +  5     or     —5. 
Hence, 

a?=4+5=9     or     a;=4— 5  =  — 1. 

That  is,  in  this  form,  the  largest  root  is  positive  and  the 
smaller  negative. 


EQUATIONS    OF    THE    SECOND    DEGREE.  183 

Verification, 

If  we  take  the  positive  value  of  x,  the  equation 
x^—Qx—^     gives     (9)2—8x9=81  —  72=9; 
and  if  we  take  the  negative  vakie,  the  equation 

a;2_-8a;=9     gives     {-l)2-8(-l)  =  l+8=9 ; 

from  which  we  see  that  both  values  alike  satisfy  the  equa- 
tion. 

2.  What  are  the  values  of  x  in  the  equation 

By  clearing  the  fractions,  we  have 

6a;2+4a;— 180  =  3a;2+12a?— 177 
and  by  transposing  and  reducing 

3a;2_8a;=3, 
and  dividing  by  the  co-efRcient  of  a;^,  we  obtain 

X^ —X=z\, 

Then,  by  completing  the  square,  we  have 

^3'^"^9~'^9       9' 

and  by  extracting  the  square  root, 

_±____^      /25 ^_5_  ___5_ 

'^'~T~~\/  9""'^T  ^^    ""3' 


Hence, 


4,5        .  ^  451 

3       3  *  3       3  3 


184  FIRST  LESSONS  IN  ALGEBRA. 

Verification, 
For  the  positive  value  of  x,  the  equation 

x^ — —x=\ 

8 
gives  32__x  3=9-8  =  1: 

and  for  the  negative  value,  the  equation 

2       Q         1 
x^ — —x=l 

o 


V      3  7       3  ^■"3~"9""^"9— ^• 


1  \2      8 
gives 


3.  What  are  the  values  of  a?  in  the  equation 

Clearing  the  fractions,  and  dividing  by  the  coefficient  of 
«2,  we  have 

Completing  the  square,  we  have 

''3''^  9 -^^+9 -36' 
then,  by  extracting  the  square  root,  we  have 


1/^7  7 

hence, 


*-T==^V36=+-6-  ""'  --e- 


1,79,,  17  5 


EQUATIONS    OF    THE    SECOND   DEGREE.  185 

Verification, 
If  we  take  the  positive  value  of  x,  the  equation 

2         2  11 

gives  (ii)2_|.Xli=2l-l  =  lJ: 

and  for  the  negative  value,  the  equation 
ac2 — jx=^H 
/      5\2      2  5      25  ,10     45     ,, 

4.  What  are  the  values  of  x  in  the  equation 
4a2_2a;2+2aa;=18a5  — 18J2  ? 

By  transposing,  changing  the  signs,  and  dividing  by  2,  it 
becomes 

x^—ax=2a^—^ab+W^  ; 

whence,  completing  the  square, 

x'^—ax+—=^—^ 9a^+962  ; 

4  4 

extracting  the  square  root. 


x=-^±:  ^^~9a5-W 

Now,  the  square  root  of    -- — Qdb+W,  is  evidently 
^—35.     Therefore, 

2      \  2  /  (  «j=r  —  a+3J. 

16* 


186  FIRST  LESSONS  IN  ALGEBRA. 

What  will  be  the  numerical  values  of  oc,  if  we  suppose 
fl=6  and  5==!  ? 


t       5.  What  are  the  values  of  x  in  the  equation 

1  4 

3  5 


—aj— 4— a;2+2a; — —x'^=:45  —  3x^-^4x  1 


,  (  a;=     7,12  )  to  within 

ix=z-5,73y       0,01. 

6.  What  are  the  values  of  x  in  the  equation 

8a;2-.14a:+10=:2a?+34  ? 

.  ( x=     3. 

Ans,     < 

<a:=  — 1. 

7.  What  are  the  values  of  x  in  the  equation 

x^ 

30+a:=2a?— 22  ? 

4 

Ans.     <     ~^ 

(x=:-4. 

8.  What  are  the  values  of  x  in  the  equation 

x2^3x+^=9x+l3i  1 
2 

Ans.     <     "" 

(x=-l. 

9.  What  are  the  values  of  x  in  the  equation 

2ax--x^z=—2ah-b^  ? 

f  x=z—b, 
10.  What  are  the  values  of  a?  in  the  equation 

a2  +  b^^2bx+x^='^1 


Ans,   < 


vz=  — —{  bn  -\-  V  ci^m^-{-b'^m'^ — o^ti^  ). 


EQUATIONS    OF    THE    SECOND    DEGREE.  187 

EXAMPLES  IN  THE  THIRD  FORM. 

1.  What  are  the  values  of  x  in  the  equation 

a:2+4xz=:— 3  ? 
First,  by  completing  the  square,  we  have 
a:2-f-4x+4=ir— 3  +  4  =  1  ; 
and  by  extracting  the  square  root, 

a:4-2=±VT=  +  l     or     —1: 
hence,     a:=:— 2  +  1  =  — 1  ;    or    0:=— 2  — 1  =  — 3. 
That  is,  in  this  form  both  the  roots  are  negative* 

Verification. 
If  we  take  the  first  negative  value,  the  equation 

gives  (_i)2+4(_i)==i_4=-3  ; 

and  by  taking  the  second  value,  the  equation 

a:2+4a:=— 3 
gives  (-3)2+4(-3)=.9-12  =  -3: 

hence,  both  values  of  x  satisfy  the  given  equation. 
2.  What  are  the  values  of  x  in  the  equation 

~^-5a:-16  =  ]2  +  -^a:2+6ir. 
2  Z 

By  transposing  and  reducing,  we  have 

—x'^—nx=2S\ 

then  since  the  coefficient  of  the  second  power  of  x  is  nega- 
tive, we  change  the  signs  of  all  the  terms  which  gives 

1^2+ 11a;  =r— 28, 


188  FIRST  LESSONS  IN  ALGEBRA* 

then  by  completing  the  square 

a;2+lla;+30,25=2,25, 
hence, 

a:— 5,5=±v^2;25^  +  l,5     or     —1,5. 
consequently, 

0?=— 4     or     xz=z—l, 

3.  What  are  the  values  of  x  in  the  equation 


— 2a?— 5=— a:2+5a:+5. 

8  8 


Ans.   \ 

i  X=z  — 


5. 


4.  What  are  the  values  of  x  in  the  equation 


2 
2aj2+8a?=— 2| — -x. 
o 


C  0:=— 4 
Ans.  s 


.  oc 
5.  What  are  the  values  of  x  in  the  equation 


3 

4a;2+— a?+3a;=:— 14a;— 3^— 4a;2. 


Ans,  ^ 


6.  What  are  the  values  of  x  in  the  equation 

3         4a;^ 
^  V  — a?2_4 — -x='-^  +24a;4-2. 

(  X=z-^l. 

7.  What  are  the  values  of  x  in  the  equation 

-i-a:2-h7a;+20=~— a;2-lla?-60. 

9  47 


Ans, 


ia;=-10. 


EQUATIONS  OF  THE  SECOND  DEGREE.      189 


8.  What  are  the  values  of  x  in  the  equation 

-—x^—xA =  — 9-i-ic — —x^ . 

6  2  ^6  2 

-8 


(  Xzzi—S 


Ans.   \ 

\  x=. — 1 


Ans. 
9.  What  are  the  values  of  x  in  the  equation 

:-10 

1 0.  What  are  the  values  of  x  in  the  equation 

x—x'^—3=6x+l. 

Ans.  <     ~~ 
\x=-l. 

1 1 .  What  are  the  values  of  x  in  the  equation 

a;2_f_4a:_90z=— 93. 

Ans.i'^^-^ 
la;=-l. 

EXAMPLES  IN  THE  FOURTH  FORM. 

1 .  What  are  the  values  of  x  in  the  equation 

x-—8x=z  —  7. 

By  completing  the  square  we  have 

a:2_8a:+16=— 7+16=:9  ; 

then  by  extracting  the  square  root 

a:— 4=dz-/9^  +  3     or     —3; 
hsnce, 

x=-{-7     or     a;=:+l- 

That  is,  in  thi«  form,  both  the  roots  are  positive. 


190  FIRST  LESSONS  IN  ALGEBEA 

Verification, 
If  we  take  the  largest  root,  tlie  equation 

a;2— 80;=— 7     gives     T^  — 8  x  7=: 49— 56=— 7; 
and  for  the  smaller,  the  equation 

a:2~8a:=— 7     gives     12  — 8  X  1=1— 8  =  — 7: 
hence,  both  of  the  roots  will  satisfy  the  equation. 

2.  What  are  the  values  of  x  in  the  equation 

40 

-l-^a:2+3a;-10  =  lla;2_i8a;+— . 

At 

By  clearing  the  fractions,  we  have 

—  3a;2+6a;— 20  =  3a;2_35^^40; 

then  by  collecting  the  like  terms 

—  6a:2+42a?=60  ; 

then  by  dividing  by  the  coefficient  of  x^,  and  at  the  same 
time  changing  the  signs  of  all  the  terms,  we  have 

0^2— 7a^=  — 10. 

By  completing  the  square,  we  have 

a:2__  7^^+12,25=2,25, 

and  by  extracting  the  square  root  of  both  members, 

a;— 3,5=±'v/2^25^  +  l,5     or     —1,5. 
hence,  ^ 

a:=3,5+l,5=5,     or     a:=3,5  — 1,5=2. 


equaTlons  of  the  second  degree.         191 

Verification. 
If  we  take  the  larger  root,  the  equation 
x^—'lx^  —  lO     gives     52— 7x5=25— 35  =  — 10; 
and  if  we  take  the  smaller  root,  the  equation 

x^—'lxzzz  —  lO     gives     22— 7x2=4— 14  =  — 10. 
3.  What  are  the  values  of  x  in  the  equation 
—  3a?+2a;2+lr:.17|a;— 2a;2— 3. 
By  transposing  and  collecting  the  terms,  we  have 

4a:2_20|a?=— 4; 
then  dividing  by  the  coefficient  of  a;^  we  have 


X' 


,2 


5ia;=.-l. 


^5 


By  completing  the  square,  we  obtain 

"=      ^^''+25-      ^+25-25' 
and  by  extracting  the  root 


2f=±\A 


2     03      ^     /  144      ^12  12 


hence, 


12     «  ^3     12      1 

.-=5;     or,     a.=2f--=- 


Verification, 
If  we  take  the  larger  root,  the  equation 
a;2— 5la?=— 1     gives     5^- 5ix5=25— 26  =  — 1 ; 
and  if  we  take  the  smaller  root,  the  equation 

(1  \  2  1       1      26 


192  FIRST  LESSONS  IN  ALGEBRA. 


4.  What  are  the  values  of  x  in  the  equation 

(ir=3. 


—x^-3x+—=  __aj2+— ^-—  ? 


Ans. 


5.  What  are  the  values  of  x  in  the  equation 
1 

T 


—  4x^ — —x-\-l^=i—'5x'^+SP  ^ 


i  X=:S, 

"    \x=l. 


Ans. 

iX:=j, 

6.  What  are  the  values  of  x  in  the  equation 

—  4:X^-\-:r:zX—-—z=:—3x^——r-X-i — -1 

20       40  20   ^40 

A71S.      <       ~*' 

7.  What  are  the  values  of  x  in  the  equation 

x^—lOj\x=z-ll 


Ans.   < 


8.  What  are  the  values  of  x  in  the  equation 

_27a:H r 1-100=-— — +12a:-26  ? 


a:=10. 

1 
TTT- 


\x=z7. 

\x=e. 

9.  What  are  the  values  of  x  m  the  equation 


Ans.   < 


-— 22a:+15= — |-28a:-30? 


Ans.   < 

(  x=l. 


10.  What  are  the  values  of  x  in  the  equation 

3^ 

"To 


2x^-30x+3=Z'-x^+3^qX^^1 


a:=:ll. 


^"Mx=A. 


EQUATIONS    OF   THE    SECOND   DEGREE.  193 

Properties  of  the  Roots, 

128.  We  have  thus  far,  only  explained  the  methods  of 
finding  the  roots  of  an  equation  of  the  second  degree.  We 
are  now  going  to  show  some  of  the  properties  of  these  roots. 

The  first  form. 

129.  The  first  form 


gives  1st   root  a?=i—  p+^/q-\-p^, 

2nd  root  xz=i~  p—^/q^p^, 

and  their  sum  =  —  2p. 

Since,  in  this  form  q  is  supposed  positive,  the  quantity 
q-\-p'^  under  the  radical  sign  will  be  greater  than  p'^,  and 
hence  its  root  will  be  greater  than  p.  Consequently  the 
first  root,  which  is  equal  to  the  difference  between  p  and 
the  radical,  will  be  positive  and  less  than  p.  In  the  second 
root,  p  and  the  radical  have  the  same  sign ;  hence,  the 
second  root  will  be  equal  to  their  sum  and  negative.  If  we 
multiply  the  two  roots  together,  we  have 

—p  +   ^/q+p'^ 


Product  equal  to —q. 


Quest. — 129.  In  the  first  form,  have  the  roots  the  same  or  contrary 
signs  1  What  is  the  sign  of  the  first  root '?  What  of  the  second  1 
Which  is  the  greater  1  What  is  their  sum  equal  to  1  What  is  their 
product  equal-  to  1 

17 


194  FIRST  LESSONS  IN  ALGEBRA, 

Hence  we  conclude, 

1  St.  That  in  the  first  form  one  of  the  roots  is  always  posi- 
tive, and  the  other  negative. 

2nd.  That  the  positive  root  is  numerically  less  than  the 
negative, 

3rd.  That  the  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  X  in  the  second  term,  taken  with  a  contrary  sign. 

4th.  That  the  product  of  the  two  roots  is  equal  to  the  known 
term  in  the  second  member,  taken  with  a  contrary  sign. 

EXAMPLES. 

1.  In  the  equation 

a?2+ir=:20, 

we  find  the  roots  to  be  4  and  —5.     Their  sum  is   —1, 
and  their  product  — 20. 

2.  In  the  equation 

a;2+2a?=:3, 

we  find  the  roots  to  be  1  and  —3.     Their  sum  is  equal  to 
—2,  and  their  product  to  —3. 

3.  The  roots  of  the  equation 

are  +9  and  —10.     Their  sum  is  —1,  and  their  product 
-90. 

4.  The  roots  of  the  equation 

are  6  and  —10.     Their  sum  is  —4,  and  their  product  is 
-60. 


EQUATIONS    OF    THE    SECOND    DEGREE.  195 

Let  these  principles  be  applied  to  each  of  the  examples 
under  "  examples  in  the  first  form." 


Second  Form. 
130.  The  second  form  is, 

and  by  resolving  the  equation  we  find 

1st  root,  oc=+p+^q-{-p^ 


2nd  root,  x  =  -{-p  —  ^q  +p^ 

and  their  sum  =  2p. 

In  this  form,  the  first  root  is  positive  and  the  second 
negative.     If  we  multiply  the  two  roots  together,  we  have 

Hence  we  conclude, 

1  St.   That  in  the  second  form  one  of  the  roots  is  positive 
and  the  other  negatwe. 

2nd.   That  the  positive  root  is  numerically  greater  than  the 
negative. 

3rd.   That  the  sum  of  the  roots  is  equal  to  the  coefficient  of 
X  in  the  second  term,  taken  with  a  contrary  sign, 

4th.   That  the  product  of  the  roots  is  equal  to  the  known 
term  in  the  second  member,  taken  with  a  contrary  sign. 


Quest. — 130.  What  is  the  sign  of  the  first  root  in  the  second  form  1 
What  is  the  sign  of  the  second  1  Which  is  the  greater  1  What  is  their 
sum  equal  to  1     What  is  their  product  equal  to  1 


IM  FIRST    LESSONS    IN    ALGEBRA. 

EXAMPLES. 

1.  The  roots  of  the  equation 

are    +4  and  —3.     Their  sum  is   -j-l,  and  their  product 
—  12. 

2.  The  roots  of  the  equation 

are  +10  and  —jt:-     Their  sum  is  9  j^,  and  their  product 
is  —1. 

3.  The  roots  of  the  equation 

£c2— 6a?=16, 

are  +8  and  —2.     Their  sum  is  +6,  and  their  product 
is   -16. 

4.  The  roots  of  the  equation 

^2— lla;  =  80, 

are  +16  and  —5.     Their  sum  is  + 1 1 ,  and  their  product 
is   —80. 

Let  these  principles  be  applied  to  each  of  the  examples 
under  "  examples  in  the  second  form." 

Third  Form.  ' 

131.  The  third  form  is, 

and  by  resolving  the  equation  we  find, 

1st  root,  x=.~p-^^J  —q-^-p^^ 


2nd  root,  x  =  —p — y— §'  -\-p^ 

Their  sum  is  =~-2p 


EQUATIONS    OP   THE    SECOND   DEGREE.  197 

In  this  form,  the  quantity  under  the  radical  being  less 
than  "p^,  its  root  will  be  less  than  p  :  hence  both  the  roots 
will  be  negative,  and  the  first  will  be  numerically  the  least. 

If  we  multiply  the  roots  together,  we  have 

{—p+ V —9-^p^)  x(—p—V —^'^P'^)=+^' 

Hence  we  conclude, 

1  St.   That  in  the  third  form  both  the  roots  are  negative. 

2nd.   That  the  first  root  is  numerically  less  than  the  second. 

3rd.  That  the  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  X  in  the  second  term,  taken  with  a  contrary  sign. 

4th.  That  the  product  of  the  roots  is  equal  to  the  known 
term  in  the  second  member,  taken  with  a  contrary  sign. 

EXAMPLES. 

1.  The  roots  of  the  equation 

are  —4  and  —5.     Their  sum  is  —9,  and  their  product 
4-20. 

2.  The  roots  of  the  equation 

a:2_^13a:=— 42, 

are   —6  and  —7.     Their  sum  is   -—13,  and  their  product 

+  42. 


Quest. — 131.  In  the  third  form,  what  are  the  signs  of  the  roots  1 
Which  root  is  the  least  \  What  is  the  sum  of  the  roots  equal  to  1 
What  is  their  product  equal  to  1 

17* 


196  FIRST    LESSONS    IN    ALGEBRA. 

3.  The  roots  of  the  equation 

3 

are and  —2.     Their  snm  is  —2f,  and  their  product 

4.  The  roots  of  the  equation 

x'^-\-5x=: — 6, 

are  —2  and  —3.      Their  sum  is   —5,  and  their  product 
is  +6  .  . 

Let  these  principles  be  applied  to  each  of  the  examples 
imder  "  examples  in  the  third  form." 

Fourth  Form, 

132.  The  fourth  form  is, 

x"^ — 2px=z — q  ; 
and  by  resolving  the  equation  we  find, 
1st  root,  x=:p-\-'^—q-^p^ 

2nd  root,  x=:p  —  ^—q-\-p^ 

Their  sum  is  =2p. 

In  this  form,  as  well  as  in  the  third,  the  quantity  under 
the  radical  being  less  than  p^,  its  root  will  be  less  than  p  : 
hence  both  the  roots  will  be  positive,  and  the  first  will  be 
the  greatest. 

If  we  multiply  the  two  roots  together,  we  have 


(p+V-9+P')x(p-V-9^P')  =  +9' 


EQUATIONS    OP   THE    SECOND   DEGREE  199 

Hence  we  conclude, 

1  St.   That  in  the  fourth  form  both  the  roots  are  positive. 

2ncl.   That  the  first  root  is  greater  than  the  second. 

3rd.  That  the  sum  of  the  roots  is  equal  to  the  coejficient  of 
X  in  the  second  term,  taken  with  a  contrary  sign. 

4th.  That  the  product  of  the  roots  is  equal  to  the  known 
term  in  the  second  member,  taken  with  a  contrary  sign. 

EXAMPLES. 

1 .  The  roots  of  the  equation 

are    +4    and    +3.     Their   sum  is    +7   and   their   pro- 
duct   +  12. 

2.  The  roots  of  the  equation 

a:2— 14ir=-24, 

are    +12    and    +2.     Their  sum  is    +14    and  their  pro- 
duct   +  24. 

3.  The  roots  of  the  equation 

a;2— 20a:=— 36, 

are    +18    and   +2.     Their  sum  is    +20    and  their  pro- 
duct   +  36. 

4.  The  roots  of  the  equation 

are    +14   and    +3.     Their  sum  is   +17    and   their  pro- 
duct   +  42. 


Quest. — 133.  In  the  fourth  form,  what  are  the  signs  of  the  roots  1 
Which  root  is  the  greatest  1  What  is  the  sum  of  the  roots  equal  to  '^ 
What  is  thfeir  product  equal  to  1 


200  PIRST    LESSONS    IN    ALGEBRA. 

133.  In  the  third  and  fourth  forms  the  values  of  x  some- 
times become  imaginary,  and  in  such  cases  it  is  necessary 
to  know  how  the  resuUs  are  to  be  interpreted. 

If  we  have  q^p^,  that  is,  if  the  known  term  is  greater 
than  half  the  coefficient  ofji  squared,  it  is  plain  that  y — 5'+^?" 
will  be  imaginary,  since  the  quantity  under  the  radical 
will  be  negative.  Under  this  supposition  the  values  of  x 
in  the  third  and  fourth  forms  will  be  imaginary. 

We  will  now  show^that,  when  in  the  third  and  fourth 
forms,  we  have  $'>p^,  the  conditions  of  the  question  will  be 
incompatible  with  each  other. 

134.  Before  showing  this  it  will  be  necessary  to  estab- 
lish a  proposition  on  which  it  depends  :  viz. 

If  a  given  number  he  decomposed  into  two  parts  and  those 
parts  multiplied  together,  the  product  will  he  the  greatest  pos- 
sible when  the  parts  are  equal. 

Let  2p  be  the  number  to  be  decomposed,  and  d  the  differ- 
ence of  the  parts.     Then 

p  -\ —         the  greater  part  (page  80,  Ex.  7.) 

and        p =         the  less  part. 

and         p2 — F~^'     their  product  (Art.  40.) 

Now  it  is  plain  that  P  will  increase  as  d  diminishes,  and 
that  it  will  be  the  greatest  possible  when    d—0  :    that  is, 

py^pz^p^     is  the  greatest  product. 


Quest. — 133.  In  which  forms  do  the  values  of  x  become  imaginary  1 
When  will  the  values  of  x  be  imaginary  1  Why  will  the  values  of  x  be 
then  imaginary  T 


or  THE  \ 

UNIVERSITY    1 

1} 


EQUATIONS  OF  THE  SECOND  DEGREE.      201 


Now,  since  in  the  equation 

Xp  is  the  sum  of  the  roots,  and  q  their  product,  it  follows 
that  q  can  never  be  greater  than  'p^.  The  conditions  of  the 
equation,  therefore,  fix  a  limit  to  the  value  of  §',  and  if  we 
make  ^>p^,  we  express  by  the  equation  a  condition  which 
cannot  be  fulfilled,  and,  this  contradiction  is  made  apparent 
by  the  values  of  x  becoming  imaginary.  Hence  we  may 
conclude  that. 

When  the  values  of  the  unknown  quantity  are  imaginary, 
the  conditions  of  the  question  are  incompatible  with  each  other, 

EXAMPLES. 

1.  Find  two  numbers  whose  sum  shall  be  12  and  pro- 
duct 46. 

Let  X  and  y  be  the  numbers. 

By  the  1st  condition,  £c+y=12  ; 

and  by  the  2d,  xy=:AQ. 

The  first  equation  gives 

x—\2—y. 
Substituting  this  value  for  x  in  the  second,  we  have 

12y— y2  — 46; 
and  changing  the  signs  of  the  terms,  we  have 
/— 12y— .— 46. 


Quest. — 134.  What  is  the  proposition  demonstrated  in  Article  1341 
If  the  conditions  of  the  question  are  incompatible,  how  will  the  values 
of  the  unknown  quantity  be  1 


202         FIRST  LESSONS  IN  ALGEBRA. 

Then  by  completing  the  square 

y2_l2y+36  = -46  +  36= -10 

which  gives  y = 6  +  -y/— 10, 

and  y  =  6— V  — 10; 

both  of  which  values  are  imaginary,  as  indeed  they  should 
be,  since  the  conditions  are  incompatible. 

2.  The  sum  of  two  numbers  is  8,  and  their  product  20  : 
what  are  the  numbers  ? 

Denote  the  numbers  by  x  and  y. 
By  the  first  condition, 

x+y=S\ 
and  by  the  second,  xy=:20. 

The  first  equation  gives 

a;=8 — y 
Substituting  this  value  of  x  in  the  second,  we  have 
8y-y2=20  ; 
changing  the  signs,  and  completing  the  square,  we  have 

y2-8y+16  =  — 4  ; 
and  by  extracting  the  root, 

y__.4_|_yCr4    and    y=:4  — -y/  — 4. 

These  values  of  y  maybe  put  under  the  forms  (Art.  106), 

y=4+2y  — 1    and   y=A:—2^/^^. 

3.  What  are  the  values  of  x  in  the  equation 

a?2+2a?=  — 10. 

Ans.    (— l  +  3/^i: 
(a?=:  — 1-3V^^1. 


EQUATIONS    OF    THE    SECOND    DEGREE.  203 

Examples  with  more  than  one  unknown  quantity. 

c  X  4-2/  —    14  ) 
1.  Given     \    ^     \     .^^  I"     to  find   x   and   y. 

By  transposing  y   in  the  first  equation,  we  have 
a;==:14 — y  ; 
and  by  squaring  both  members, 

Substituting  this  value  for  x^  in  the  2nd  equation,  we  have 
196-28y+y2+y2_ioO; 

from  which  we  have 

y2_l4y=— 48; 
and  by  completing  the  square, 

y2_14^_[_49_l    ; 

and  by  extracting  the  square  root, 

y-7=:d=VT=r+l  or  -1: 
hence,  y=:7+l=:8,    or    y=:7  — 1=6. 

If  we  take  the  larger  value,  we  find  xz=zQ  \  and  if  we 
take  the  smaller,  we  find  a?  =  8. 

Verification. 
For  the  largest  value,  y=8,  the  equation 
a:-f-y=:14     gives     6  +  8r=14; 
and  a:2-fy2_ioo     gives     36  +  64  =  100. 

For  the  value  y  =  6,  the  equation 

a;4-y=14     gives     8  +  6  =  14  ; 
and  a;2+y2=ioo     gives     64  +  36  =  100. 

Hence,  both  sets  of  values  will  satisfy  the  given  equation. 


204  FIRST    LESSONS    IN    ALGEBRA. 

2.  Given     {    '      \,~~_  >     to  find  x  and  y. 

(  a;2— y2__45  )  ^ 

Transposing  y  in  the  first  equation,  we  have 

a?=3+y  ; 

and  then  squaring  both  members, 

x"z=:z9-{-6i/-\-y'^. 

Substituting  this  value  for  x"^  in  the  second  equation,  we 
have 

9  +  6y+y2 — y2__45  . 

whence  we  have 

6yz=:36    and    yr=6. 

Substituting  this  value  of  y  in  the  first  equation,  we  hav^ 

a?— 6  =  3, 

and  consequently  x=3  +  6=z9. 

Verification. 
x—y=z3     gives     9  —  6  =  3  ; 
and  ir2— 3/2  —  45     gives     81—36=45. 

,.     ^.  ^  a;24-3^y  =22  ) 

3.  Given      {    ^  ,  ^       ,  ^  n      .r^\      to  find  x  and  y, 

I  a;2-f-3a:y+ 21/2  —  40  )  ^ 

Subtracting  the  first  equation  from  the  second,  we  have 
2y2  =  18, 
which  gives  y^  =  9, 

and  y=-}-3    or    —3. 

Substituting  the  plus  value  in  the  first  equation,  we  have 
a:2+9a;=22; 


EQUATIONS    OF    THE    SECOND    DEGREE.  205 

from  wh^ch  we  find 

x—-{-2     and     ic— — 11. 

If  we  take  the  negative  value,  y=r:— 3,  we  have  from  the 
first  equation, 

0^2—9^—22  ; 

from  which  we  find 

iczz:  +  ll     and     x=—2. 

Verification. 

For  the  values  y=r:+3  and  x=:+2,  the  equation 

a:2+3a:y=22 

gives  22+3x2x3  =  4+18=22: 

and  for  the  second  value,  a:  =—11,  the  same  equation 

a:2  +  3a:y=22 

gives         (  —  ll)2  +  3x  -11X3  =  121— 99=22. 

If  now  we  take  the  second  value  of  y,  that  is,  y— —  3, 
and  the  corresponding  values  of  x,  viz,  a:=  +  ll,  and 
a:=— 2;    for  a:=  +  ll,  the  equation 

ir2+3a:y=22 
gives  112+3x11  X-3  =  121-99=22  ; 

and  for  x=z—2,  the  same  equation 

5c2  +  3a:y=22 
gives  (-2)2+3x -2  X -3=4+18=22. 


r  xz=f     (l)x 

Jiven  <  X  +y  +z  =7     (2)  >  to  find  x,- 
(  a;2  I  ^2_j_^2_2i      (3)  ) 


xz=zy'^  (1) 

4.  Given  ^  x  +y  +z  =7  (2)  ^  to  find  x,i/,  and 

a;2+y2_j_^2_2i  (3) 
18 

I 


206  FIRST  LESSONS  IN  ALGEBRA. 

Transposing  y   in  the  second  equation,  we  have 

x+z^l-y    (4)  ; 

then  squaring  the  members,  we  have 

a:2+2a:;^+;^2_49_i4y_|_y2. 

If  now  we  substitute  for    2xz   its  value  taken  from  the 
first  equation,  we  have 

a;2+2y24-;^23^49„14y^y2  . 

and  cancelKng  y^  ij^  each  member,  there  results 
a:2+y2+;^2_49_i4y. 

But,  from  the  third  equation  we  see  that  each  member  of 
the  last  equation  is  equal  to  21  :  hence 

49-14y=21, 

and  14y=i49— 21rz:28. 

28     ^ 
hence,  y=:— -=:3. 

^      14 

Placing  this  value  for  y  in  equation  (1)  gives 

xzz=L^  ; 

and  placing  it  in  equation  (4)  gives 

x-\-z:=zb,     and     a:z=5  — 2-. 

Substituting  this  value  of  x  in  the  previous  equation,  we 
obtain 

52^— >^2_4       Qj.       2:2_52:— _4j 

and  by  completing  the  square,  we  have 
^2_5^_|.6^25=2,5, 
and  £r— 2,5r=±V2^=-{-l>5     or     —1,5; 

hence,       ^=2,5  +  1,5=4     or     2^=4-2,5  —  1,5=1. 


EQUATIONS    OF    THE    SECOND    DEGREE.  207 

If  we  take  the  value 

z=:4,     we  find     a?=l  : 
if  we  take  the  less  value 

z=:},     we  find     ocz=4. 

t  3.  Given     x  +^/ocy+y  =   19  >      ^    r    i  i 

'    y    y  '  ^  >      to  find  X  and  y. 

and        x^+     x7/-\-7/=:l33  )  ^ 

Dividing  the  second  equation  by  the  first,  we  have 

x—'\/^-\-  y=  7 

but  a:+'\/^+   y=19 

hence,  by  addition,  2x  +2y=26 

or  a?  +  y=:13 
and  substituting  in  1  st  equa.     '\/Ty+ 13  =  19 

or  -y/xy^zz  6 

and  by  squaring  a?y=36 

From  2d  equation,  x'^+xy +y'^  =  l33 

and  from  the  last  3xy        =108 

Subtracting  x'^—2xy-\-y'^z:=  25 

hence,  x—y:=ziz     5 

but  a?+y=:        13 

hence               a?=9  or   4 ;     and   y=4    or    9. 

6.  Given  the  s«um  of  two  numbers  equal  to  a,  and  the 
sum  of  their  cubes  equal  to  c,  to  find  the  numbers 


By  the  conditions  <    „     ^, 


208  FIRST  LESSONS  IN  ALGEBRA. 

Putting         x=:s-{-z,     and     y—s—z,     we  have 
a=2s,     or     5=—; 

\  7/-s^  —  3s^z+3sz^—z^, 
hence,  by  addition,    x^-\-y''^=2s^  -\-6sz'^z=zc, 

whence  z^=z — - —     and     z=dz\/ — ; 

OS  \  OS 


r^zz2s'3  7 

a?=:5±Y/— ^^;     and     y=s^i^^ 


c—2s^ 


6s 

or  by  putting  for  s  its  vakie, 

a3  

'=-2-V  V-37-)=T-V-l27-' 


and  y==-J=Fy/(i^)=.|-^y/i^ 


2a    ' 

Note. — What  are  the  numbers  when  a  — 5  and  c=r35. 
What  are  the  numbers  when   az=i9    and   c  =  243. 

QUESTIONS. 

1.  Find  a  number  such,  that  twice  its  square,  added  to 
three  times  the  number,  shall  give  65. 

Let  X  denote  the  unknown  number.  Then  the  equation 
of  the  problem  will  be 

2x'^+3x.=z65  , 
whence 


EQUATIONS    OF    THE    SECOND    DEGREE.  209 

Therefore, 

3    ,  23     ^  ^  3      23  13 

Both  these  values  satisfy  the  question  in  its  algebraic 
sense.     For, 

2x(5)2+3x5=2x25  +  15r=:65  ; 
13\2     ^  13      169      39      130 


and 


/      13V  .  o  13       169       39       13U       ^^ 


Remark. — If  we  wish  to  restrict  the  enunciation  to  its 
arithmetical  sense,  we  will  first  observe,  that  when  x  is 
replaced  by  —x,  in  the  equation  2x'^-\-3x=z65j  the  sign  of 
the  second  term  3x  only,  is  changed,  because  (—xY=x^. 

3      23 

Therefore,  instead  of  obtaining  a:= — — ±— ,  we  should 

find    xz= — dt — ,  or  xz= — ,  and  a:  =—5,  values  which  only 
4       4  2 

differ  from  the  preceding  by  their  signs.     Hence,  we  may 

13 

say  tliat  the  negative  solution ,    considered    indepen- 

dently  of  its  sign,  satisfies  this  new  enunciation,  viz  :  To 
find  a  numher  such,  that  twice  its  square,  diminished  hy  three 
times  the  numher,  shall  give  65.     In  fact,  we  have 


«      /13\2     ^     13      169      39     ^ 


Remark. — The  root  which  results  from  giving  the  plus 
sign  to  the  radical,  generally  resolves  the  question  both 
in  its  arithmetical  and  algebraic  sense,  while  the  second 
root  resolves  it  in  its  algebraic  sense  only. 

18* 


210  FIRST  LESSONS  IN  ALGEBRA. 

Thus,  in  the  example,  it  was  required  to  find  a  number, 
of  which  twice  the  square  added  to  three  times  the  mmiber 
shall  give  Qb.  Now,  in  the  arithmetical  sense,  added  means 
increased  ;  but  in  the  algebraic  sense  it  implies  diminution, 
when  the  quantity  added  is  negative.  In  this  sense,  the 
second  root  satisfies  the  enunciation. 

2.  k  certain  person  purchased  a  number  of  yards  of  cloth 
for  240  cents.  If  he  had  received  3  yards  less  of  the  same 
cloth  for  the  same  sum,  it  would  have  cost  him  4  cents  more 
per  yard.     How  many  yards  did  he  purchase  ? 

Let         0?=       the  number  of  yards  purchased. 

Then     will  express  the  price  per  yard. 

If,  for  240  cents,  he  had  received  3  yards  less,  that  is 
a?— 3  yards,  the  price  per  yard,  under  this  hypothesis,  would 

this  last  cost  would  exceed  the  first  by  4  cents.     Therefore, 
we  have  the  equation 

240       240      , 
=4; 

X  —  6         X 

whence,  by  reducing    a?^  — 3a; =180, 


3  /W    ,^^       3=t27 


2       ' 
therefore  x=l6     and     x=:  —  \2. 

The  value  a:=15  satisfies  the  enunciation;  for,  15  yards 

24-0 
for  240  cents  gives  ,     or  16  cents  for  the  price  of 

1  o 

one  yard,  and  12  yards  for  240  cents,  gives  20  cents  forth© 

price  of  one  yard,  which  exceeds  16  by  4. 


EQUATIONS    OF    THE    SECOND    DEGREE.  211 

As  to  the  second  solution,  we  can  form  a  new  enuncia- 
tion, with  which  it  will  agree.  For,  going  back  to  the 
equation,  and  changing  x  into  —a?,  it  becomes 

240         240       ,               240       240       , 
-=4,     or r^=4, 


— X — 3       — X  X        jr+3 

an  equation  which  may  be  considered  the  algebraic  transla- 
tion of  this  problem,  viz  :  A  certain  person  purchased  a  num- 
ber of  yards  of  cloth  for  240  cents :  if  he  had  paid  the  same 
sum  for  3  yards  more,  it  would  have  cost  him  4  cents  less  per 
yard.     How  many  yards  did  he  purchase  ? 

Ans.  a;  =  12,  and  a?=  — 15. 

3.  A  man  bought  a  horse,  which  he  sold  after  some  time 
for  24  dollars.  At  this  sale,  he  loses  as  much  per  cent, 
upon  the  price  of  his  purchase  as  the  horse  cost  him. 
What  did  he  pay  for  the  horse  ? 

Let  X  denote  the  number  of  dollars  that  he  paid  for  the 
horse,  a;— 24  will  express  the  loss  he  sustained.     But  as 

he  lost  X  per  cent,  by  the  sale,  he  must  have  lost    

^  -^  '  100 

upon  each  dollar,  and  upon  x  dollars  he  loses  a  sum  de- 
noted by   -jTTfr  »  ^'^  h^B.YQ  then  the  equation 

z=zx—2A,     whence     a?2  — 100^?=:— 2400. 


100 


and  a?=50zbv^2500— 2400  =  50dzl0. 

Therefore,  a? =60    and   a:=:40. 

Both  of  these  values  satisfy  the  question. 

For,  in  the  first  place,  suppose  the  man  gave  %Q0  for  the 
horse  and  sold  him  for  24,  he  loses  36.  Again,  from  the 
enunciation,  he  should  lose  60  per  cent,   of  60,   that   is, 


212  FIRST  LESSONS  IN  ALGEBRA* 

of  60,  or   —  ,    which   reduces    to    36  ;    there- 


100  '  100 

fore,  60  satisfies  the  enunciation. 

Had  he  paid  $40,  he  would  have  lost  $16  by  the  sale  ; 

40 
for,  he  should  lose  40  per  cent,  of  40,  or    40  x  — -f-^,    which 

reduces  to  16  ;  therefore,  40  verifies  the  enunciation. 

4.  A  man  being  asked  his  age,  said  the  square  root  of 
my  own  age  is  half  the  age  of  my  son,  and  the  sum  of 
our  ages  is  80  years  :  what  was  the  age  of  each  1 

Let     a?=     the  age  of  the  father. 
y  =r     that  of  the  son. 
Then  by  the  first  condition 

and  by  the  second  condition 

a:+y=80. 
If  we  take  the  first  equation 

and  square  both  members,  we  have 

If  we  transpose   y  in  the  second,  we  have 
a:=80— y: 
from  which  we  find 

y=:— 2zfc'v/324'=:16; 

by  taking  the  plus  root,  which  answers  to  the  question  in 
its  arithmetical  sense.  Substituting  this  value,  we  find 
a? =64.  ^       ^  Father's  age  64 

(  Son's  16. 


^'  EQUATIONS    OF    THE    SECOND    DEGJIEE.  213 

5.  Find  two  numbers,  such  that  the  sum  of  their  pro- 
ducts by  the  respective  numbers  a  and  J,  may  be  equal  to 
Si",  and  that  their  product  may  be  equal  to  p. 

Let  X  and  y  be  the  required  numbers,  we  have  the  equa- 
tions 

ax-\-hyz=.%s, 
and  xyz=i'p, 

2s — ax 
From  the  first  y= — ; 

whence,  by  substituting  in  the  second,  and  reducing, 
ax? — 2sx  =z  —  bp. 
1 


Therefore,  x— — ± — a/ s^—abp, 

a       a 

and  consequently, 

s       1 


y=—::pj-^S^-abp. 

This  problem  is  susceptible  of  two  direct  solutions,  be- 
cause s  is  evidently  >  ^s^—abp  ;  but  in  order  that  they 
may  be  real,  it  is  necessary  that  s^^   or   =zabp. 

Let  a  =  b—i  ;  the  values  of  x  and  y  reduce  to 

x=:sdz^s^ — p     and     y  — ^^p-y/^^ — p. 

Whence  we  see,  that  the  two  values  of  x  are  equal  to 
those  of  y,  taken  in  an  inverse  order ;  which  shows,  that  if 
s-{-  ^/ s^ — p  represents  the  value  of  a?,  s —  aJs^ — p  will  re- 
present the  corresponding  value  of  y,  and  reciprocally. 

This  circumstance  is  accounted  for,  by  observing  that  in 
this  particular  case  the  equations  reduce  to 


\     xy—p  ;  S 


214  FIRST    LESSONS    IN    ALGEBRA. 

and  then  the  question  is  reduced  to,  finding  two  numbers  of 
lohich  the  sum  is  2s,  and  their  product  p,  or  in  other  words, 
to  divide  a  number  2s,  into  two  such  parts,  that  theit  product 
may  be  equal  to  a  given  number  p. 

Let  us  now  suppose 

2sz=l4:     and    jt>  — 48: 
what  will  then  be  the  values  of  x  and  y  ? 


Ans.     \  ^ 


:8  or  6. 
'  y=6  or  8. 

6.  A  grazier  bought  as  many  sheep  as  cost  him  £Q0,  and 
after  reserving  fifteen  out  of  the  number,  he  sold  thtj  re- 
mainder for  jC54,  and  gained  2s.  a  head  on  those  he  sold  : 
how  many  did  he  buy  I  Ans.  75. 

7.  A  merchant  bought  cloth  for  which  he  paid  .£"33  156\, 
which  he  sold  again  at  £2  Qs.  per  piece,  and  gained  by  the 
bargain  as  much  as  one  piece  cost  him  :  how  many  pieces 
did  he  buy  ?  Ans.  15. 

8.  What  number  is  that,  which,  being  divided  by  the  pro- 
duct of  its  digits,  the  quotient  is  3  ;  and  if  18  be  added  to 
it,  the  digits  will  be  inverted  ?  Ans.  24. 

9.  To  find  a  number,  such  that  if  you  subtract  it  from  1 0, 
and  multiply  the  remainder  by  the  nmnber  itself,  the  product 
shall  be  21.  Ans.  7  or  3. 

10.  Two  persons,  A  and  B,  departed  from  different  places 
at  the  same  time,  and  travelled  towards  each  other.  On 
meeting,  it  appeared  that  A  had  travelled  18  miles  more 
than  B  ;  and  that  A  could  have  gone  B's  journey  in  15|^ 
days,  but  B  would  have  been  28  days  in  performing  A's 
journey.     How  far  did  each  travel  ? 

^         (  A  72  miles, 
Ans.    \ 

B  54  milea. 


EQUATIONS    OF    THE    SECOND    DEGREE.  215 

11.  There  are  two  numbers  whose  difference  is  15,  and 
half  their  product  is  equal  to  the  cube  of  the  lesser  num- 
ber.    What  are  those  numbers  ?  Ans.  3  and  18. 

12.  What  two  numbers  are  those  whose  sum,  multiplied 
by  the  greater,  is  equal  to  77  ;  and  whose  difference,  multi- 
plied by  the  lesser,  is  equal  to  12  ? 

Ans.   4  and  7,  or  |  -y/T  and  y  ^/^, 

13.  To  divide  100  into  two  such  parts,  that  the  sum  of 
their  square  roots  may  be  14.  Ans.  64  and  36. 

14.  It  is  required  to  divide  the  number  24  into  two  such 
parts,  that  their  product  may  be  equal  to  35  times  their  dif- 
ference. Ans,  10  and  14. 

15.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their 
cubes  152.     What  are  the  numbers  ?  Ans,  3  and  5. 

16.  Two  merchants  each  sold  the  same  kind  of  stuff; 
the  second  sold  3  yards  more  of  it  than  the  first,  and  to- 
gether they  receivne  35  dollars.  The  first  said  to  the  second, 
*' 1  would  have  received  24  dollars  for  your  stuff;"  the 
other  replied,  "  And  I  should  have  received  12^  dollars  for 
yours."     How  many  yards  did  each  of  them  sell  ? 

.  C  1st  merchant  a? =15  oc=b. 

Ans.     I       ,  or 

;i2nd        „       y=\Q      "^     y=8. 

17.  A  widow  possessed  13,000  dollars,  which  she  divided 
into  two  parts,  and  placed  them  at  interest,  in  such  a  man- 
ner, that  the  incomes  from  them  were  equal.  If  she  had 
put  out  the  first  portion  at  the  same  rate  as  the  second,  she 
would  have  drawn  for  this  part  360  dollars  interest ;  and  if 
she  had  placed  the  second  out  at  the  same  rate  as  the  first, 
she  would  have  drawn  for  it  490  dollars  interest.  What 
were  the  two  rates  of  interest  ? 

Ans.  7  and  6  per  cent. 


216  FIRST  LESSONS  IN  ALGEBRA. 


CHAPTER  VII. 
Of  Proportions  and  Progressions. 

135.  Two  quantities  of  the  same  kind  may  be  compared 
together  in  two  ways  : — 

1st.  By  considering  how  much  one  is  greater  or  less  than 
the  other,  which  is  shown  by  their  difference  ;  and, 

2nd.  B;y  considering  how  many  times  one  is  greater  or 
less  than  the  other,  which  is  shown  by  their  quotient. 

Thus,  in  comparing  the  numbers  3  and  12  together  with 
respect  to  their  difference,  we  find  that  ]  2  exceeds  3  by  9 ; 
and  in  comparing  them  together  with  respect  to  their  quo- 
tient, we  find  that  12  contains  3  four  times,  or  that  12  is  4 
times  as  great  as  3. 

The  first  of  these  methods  of  comparison  is  called  Arith- 
metical Proportion,  and  the  second  Geometrical  Proportion. 

Hence,  Arithmetical  Proportion  considers  the  relation  of 
quantities  with  respect  to  their  difference,  and  Geometrical 
Proportion  the  relation  of  quantities  with  respect  to  their 
quotient. 


Quest. — 135.  In  how  many  ways  may  two  quantities  be  compared 
together  1  What  does  the  first  method  consider  1  What  the  second  1 
What  is  the  first  of  these  methods  called  1  What  is  the  second  called  ? 
How  then  do  you  define  the  two  proportions  ^ 


ARITHMETICAL    PROPORTION.  217 

Of  Arithmetical  Proportion  and  Progression, 

136.  If  we  have  four  numbers,  2,  4,  8,  and  10,  of 
which  the  difference  between  the  first  and  second  is  equal 
to  the  difference  between  the  third  and  fourth,  these  num- 
bers are  said  to  be  in  arithmetical  proportion.  The  first 
term  2  is  called  an  antecedent^  and  the  second  term  4,  with 
which  it  is  compared,  a  consequent.  The  number  8  is  also 
called  an  antecedent,  and  the  number  10,  with  which  it  is 
compared,  a  consequent. 

When  the  difference  between  the  first  and  second  is  equal 
to  the  difference  between  the  third  and  fourth,  the  four  num- 
bers are  said  to  be  in  proportion.     Thus,  the  numbers 

2,     4,     8,     10, 

are  in  arithmetical  proportion. 

137.  When  the  difference  between  the  first  antecedent 
and  consequent  is  the  same  as  between  any  two  adjacent 
terms  of  the  proportion,  the  proportion  is  called  an  arith- 
metical  progression.  Hence,  a  progression  by  differences,  or 
an  arithmetical  progression,  is  a  series  in  which  the  succes- 
sive terms  continually  increase  or  decrease  by  a  constant 
number,  which  is  called  the  common  difference  of  the 
progression. 

Thus,  in  the  two  series 

1,     4,     7,  10,  13,  16,  19,  22,  25,  .  .  . 
60,  56,  52,  48,  44,  40,  36,  32,  28,  .  .  . 


Quest. — 136.  When  are  four  numbers  in  arithmetical  proportion  1 

What  is  the  first  called  1     What  is  the  second  called  1     What  is  th© 
third  called  1     What  is  the  fourth  called "? 

19 


218         FIRST  LESSONS  IN  ALGEBRA. 

the  first  is  called  an  increasing  progression,  of  which,  the 
common  difference  is  3,  and  the  second  a  decreasing  pro- 
gression, of  which  the  common  difference  is  4. 

In  general,  let  a,  b,  c,  d,  e,f,  .  .  .  designate  the  terms 
of  a  progression  by  differences  ;  it  has  been  agreed  to  write 
them  thus : 

a,h.c.d,e.f.g.h.i.h... 

This  series  is  read,  a  is  to  h,  as  h  is  to  c,  as  c  is  to  d,  as  d 
is  to  e,  Slc.  This  is  a  series  of  continued  equi-differences, 
in  which  each  term  is  at  the  same  time  a  consequent  and 
antecedent,  with  the  exception  of  the  first  term,  which  is 
only  an  antecedent,  and  the  last,  which  is  only  a  consequent. 

138.  Let  T  represent  the  common  difference  of  the 
progression 

a  *  h  .  c  ,  d  .  e  .  f  .  g  .  h,  &c, 

which  we  will  consider  increasing. 

From  the  definition  of  the  progression,  it  evidently  follows 
that 

hz:za-\-r,     c=:5  +  rr=a+2r,     d—c-\-rz=:a+Zr', 

and,  in  general,  any  term  of  the  series  is  equal  to  the  first 
term  plus  as  many  times  the  common  difference  as  there  are 
preceding  terms. 

Thus,  let  I  be  any  term,  and  n  the  number  which  marks 
the  place  of  it :  the  expression  for  this  general  term  is 

l=a-{-{n  —  \)r. 


Quest. — 137.  What  is  an  arithmetical  progression  1  What  is  the 
number  called  by  which  the  terms  are  increased  or  diminished  1  What 
is  an  increasing  progression  'I  What  is  a  decreasing  progression  '^ 
Which  term  is  only  an  antecedent  %     "WTiich  only  a  consequent  1 


f 

ARITHMETICAL    PROGRESSION.  219 

Hence,  for  finding  the  last  term,  we  have  the  following 

RULE. 

I.  Multiply  the  common  difference  by  one  less  than  the 
numher  of  terms. 

II.  To  the  product  add  the  first  term:  the  sum  will  he  the 
.last  term. 

EXAMPLES. 

The  formula  Z~a+(»  — 1)?'  serves  to  find  any  term 
whatever,  without  our  being  obliged  to  determine  all  those 
which  precede  it. 

1.  If  we  make  n=i,  we  have  /=«;  that  is,  the  series 
will  have  but  one  term. 

2.  If  we  make  ni=2,  we  have  l^^a-\-r  ;  that  is,  the  series 
will  have  two  terms,  and  the  second  term  is  equal  to  the 
first  plus  the  common  diflference. 

3.  If  a  — 2  and  r—2,  what  is  the  3rd  term?        Ans,  7. 

4.  If  «=5  and  r—A,  what  is  the  6th  term?      Ans.  25. 

5.  If  az=7  and  r=:5,  what  is  the  9th  term?      Ans.  47. 

6.  If  a=z8  and  r=5,  what  is  the  10th  term  ? 

Ans.  53. 

7.  If  a— 20  and  r—^,  what  is  the  12th  term? 

Ans.  64. 

8.  If  a—AO  and  r=z20,  what  is  the  50th  term  ? 

Ans.  1020. 


Quest. — 138.  Give  the  rule  for  finding  the  last  term  of  a  series  when 
the  progression  is  increasing. 


22©  FIRST  LESSONS  IN  ALGEBRA. 

9.  If  a=45  and  r=30,  what  is  the  40th  tenii  ? 

Ans.  1215. 
la.  If  a=z30  and  r=20,  what  is  the  60th  term? 

Ans.  1210. 

11.  If  a  =  50  and  rz=:10,  what  is  the  100th  term? 

Ans.  1040. 

12.  To  find  the  50th  term  of  the  progression 

1   .  4  .  7  .  10  .  13  .  16  .  19  .  .  ., 
we  have  /r=l 4-49x3  =  148. 

13.  To  find  the  60th  term  of  the  progression 

1   .  5  .  9  .  13  .  17  .  21   .  25  .  .  ., 
we  have  Z=  1+59x4=237. 

\        139.  If  the  progression  were   a  decreasing  one,  we 
should  have 

l=a  —  {n  —  l)r. 

Hence,  to  find  the  last  term  of  a  decreasing  progression^ 
we  have  the  following 

RULE. 

I.  Multiply  the  common  difference  hy  one  less  than  the  num-- 
her  of  terms. 

II.  Subtract  the  product  from  the  first  term  :  the  remainder 
will  be  the  last  term. 


Quest. — 139.  Give  the  rule  for  finding  the  last  term  of  a  series* 
when  the  progression  is  decreasing. 


ARITHMETICAL    PROGRESSION.  221 


EXAMPLES. 

1 .  The  first  term  of  a  decreasing  progression  is  60,  the 
number  of  terms  20,  and  the  common  difference  3  :  what 
is  tne  last  term  ? 

l~a  —  {n  —  l)r     gives     Zr:::60— (20  — 1)3  =  60  — 57=3. 

2.  The  first  term  is  90,  the  common  difference  4,  and 
the  number  of  terms  15  :  what  is  the  last  term  ?     Ans,  34. 

3.  The  first  term  is  100,  the  number  of  terms  40,  and  the 
common  difference  2  :  what  is  the  last  term  ?         Ans.  22. 

4.  The  first  term  is  80,  the  number  of  terms  10,  and  the 
common  difference  4  :  what  is  the  last  term  ?         Ans,  44. 

5.  The  first  term  is  600,  the  number  of  terms  100,  and 
the  common  difference  5  :  what  is  the  last  term  ? 

Ans.  105. 

6.  The  last  term  is  8^00,  the  number  of  terms  200,  and 
the  common  difference  2  :  what  is  the  last  term  ? 

Ans.  402. 

140.  A  progression  by  differences  being  given,  it  is 
proposed  to  prove  that,  the  sum  of  any  two  terms,  taken  at 
equal  distances  from  the  two  extremes,  is  equal  to  the  sum  of 
the  two  extremes. 

That  is,  if  we  have  the  progression 

2  .  4  .  6  .  8  .  10  .  12, 

we  wish  to  prove  that 

4+10     or     64-8 

is  equal  to  the  sum  of  the  two  extremes  2  and  12. 

19* 


222  FIRST  LESSONS  IN  ALGEBRA. 

Let  a,h,c.d,e.f,...i.k.l  be  the  pro- 
posed progression,  and  n  the  number  of  terms. 

We  will  first  observe  that,  if  x  denotes  a  term  which  has 
p  terms  before  it,  and  y  a  term  which  has  p  terms  after  it, 
we  have,  from  what  has  been  said, 

x=a-\-pXr, 

and  2/=l—px  T  ; 

whence,  by  addition,      x-\-yz=.a-\-l. 

Which  demonstrates  the  proposition. 

Referring  this  proof  to  the  previous  example,  if  we  sup- 
pose, in  the  first  place,  x  to  denote  the  second  term  4,  then 
y  will  denote  the  term  10,  next  to  the  last.  If  x  denotes 
the  3rd  term  6,  then  y  will  denote  8,  the  third  term  from 
the  last. 

Having  proved  the  first  part  of  the  proposition,  write  the 
progression  below  itself,  but  in  an  inverse  order,  viz : 

a,h.c.a.e.f.,.i.li.l. 

I  ,  k  .  i c  .  b  .  a. 

Calling  S  the  sum  of  the  terms  of  the  first  progression, 
2S  will  be  the  sum  of  the  terms  in  both  progressions,  and 
we  shall  have 

2S=:(a+l)  +  {b+k)  +  {c+i)  .  .  .  -^{i-^c)  +  {k-\-b)+{l+a). 

Now,  since  all  the  parts  a+I,  b-{-k,  c+i  .  .  .  are  equal 
to  each  other,  and  their  number  equal  to  n, 

2S=(a+l)n,     or     S=(^)n. 


ARITHMETICAL    PROGRESSION.  223 

Hence,  for  finding  the  sum  of  an  arithmetical  series,  we 
have  the  following 


RULE. 

I.  Add  the  two  extremes  together,  and  take  half  their  sum. 

II.  Multiply  the  half-sum  hy   the   number  of  terms ;    the 


product  will  he  the  sum  of  the  series. 


EXAMPLES. 


1.  The  extremes  are  2  and  16,  and  the  number  of  terms 
8  :  what  is  the  sum  of  the  series  ? 


S-- 


=:(-^jXn,     gives      Srr:— ^ x8  =  72. 


2.  The  extremes  are  3  and  27,  and  the  number  of  terms 
12  :  what  is  the  sum  of  the  series  1  Ans.  180. 

3.  The  extremes  are  4  and  20,  and  the  number  of  terms 
10  :  what  is  the  sum  of  the  series  ?  Ans.  120. 

4.  The  extremes  are  100  and  200,  and  the  number  of 
terms  80  :  what  is  the  sum  of  the  series  ?        Ans.  12000. 

5.  The  extremes  are  500  and  60,  and  the  number  of  terms 
20  :  what  is  the  sum  of  the  series  ?  Ans.  5600. 

6.  The  extremes  are  800  and  1200,  and  the  number  of 
terms  50  :  what  is  the  sum  of  the  series  ?         Ans.  50000. 


Quest. — 140.  In  every  progression,  what  is  the  sum  of  the  two  ex- 
tremes equal  to  1  What  is  the  rule  for  finding  the  sum  of  an  arithmeti 
cal  series  1 


224  FIRST  LESSONS  IN  ALGEBRA. 

>  '^       1 4 1 .  In  arithmetical  proportion  there  are  five  numbers 
to  be  considered  : — 

1st.  The  first  ^rm,  a. 

2nd.  The  common  difference,  r. 

3rd.  The  number  of  terms,  n. 

4th.  The  last  term,  Z. 

5th.  The  sum,  S. 

The  formulas 

l—a-{-[n  —  \)r     and     Sz=i(  )xn 

contain  five  quantities,  a,  r,  w,  Z,  and  S,  and  consequently 
give  rise  to  the  following  general  problem,  viz  :  Any  three 
of  these  five  quantities  being  given,  to  determine  the  other 
two. 

We  already  know  the  value  of  S  in  terms  of  a,  n,  and  r. 
From  the  formula 

l=:a-\-[n—\)r, 
we  find  a—l—{n—\)r. 

That  is  :    The  first  term  of  an  increasing  arithmetical  pro- 
gression is  equal  to  the  last  term,  minus  the  product  of  the 
common  difference  hy  the  number  of  terms  less  one. 
From  the  same  formula,  we  also  find 

I — a 

That  is  :  In  any  arithmetical  progression,  the  common  differ- 
ence is  equal  to  the  difference  between  the  two  extremes  divided 
by  the  number  of  terms  less  one. 

Quest. — 141.  How  many  numbers  are  considered  in  arithmetical 
proportion  1  What  are  they  1  In  every  arithmetical  progression,  what 
is  the  common  diiference  equal  to  1 


ARITHMETICAL    PROGRESSION.  225 

The  last  term  is  16,  the  first  term  4,  and  the  number  of 
terms  5  :  what  is  the  common  difference  ? 

The  formula  rz=z 

n—\ 

16—4      ^ 
gives  rz± —3. 


2.  The  last  term  is  22,  the  first  term  4,  and  the  number 
of  terms  10  :  what  is  the  common  difierence  ?  Ans.  2. 

142.  The  last  principle  affords  a  solution  to  the  follow- 
ing question  : 

To  find  a  number  m  of  arithmetical  means  between  two 
given  numbers  a  and  b. 

To  resolve  this  question,  it  is  first  necessary  to  find  the 
common  difference.  Now,  we  may  regard  a  as  the  first 
term  of  an  arithmetical  progression,  h  as  the  last  term,  and 
the  required  means  as  intermediate  terms.  The  number  of 
terms  of  this  progression  will  be  expressed  by  m4-2. 

Now,  by  substituting  in  the  above  formula,  h  for  Z,  and 
w-}-2  for  n^  it  becomes 

h  —  a  h  —  a 


m-}-2  — 1        m-\-\ 


that  is,  the  common  difference  of  the  required  progression  is 
obtained  by  dividing  the  difference  between  the  given  num- 
bers a  and  5,  by  one  more  than  the  required  number  of 
means. 


Quest. — 142.  How  do  you  find  any  number  of  arithmetical  means 
between  two  given  numbers  1 


226  FIRST  LESSONS  IN  ALGEBRA. 

Having  obtained  the  common  difference,  form  the  second 
term  of  the  progression,  or  the  first  arithmetical  mean^  by 

adding  r.  or   ,  to  the  first  term  a.     The  second  mean 

"^     '         m-\-\ 

is  obtained  by  augmenting  the  first  by  r,  &c. 

1.  Find  three  arithmetical  means  between  the  extremes 
2  and  18. 


The  formula 


m+1 


18-2       , 
gives  r=  — - — =4  ; 

hence,  the  progression  is 

2  .  6  .  10  .  14  .   18. 

2.  Find  twelve  arithmetical  means  between  12  and  77. 
h-a 


The  formula 


m-\-l 


77-12     ^ 
gives  r:=z — — — —5, 

Hence  the  progression  is 

12  .   17  .  22  .  27  ....  77 

143.  Remark.  If  the  same  number  of  arithmetical 
means  are  inserted  between  all  the  terms,  taken  two  and 
two,  these  terms,  and  the  arithmetical  means  united,  will 
form  but  one  and  the  same  progression. 

For,  let  a.h.c,d.e.f...  be  the  proposed 
progression,  and  m  the  number  of  means  to  be  inserted 
between  a  and  h,  h  and  c,  c  and  d  .  ,  , 


ARITHMETICAL    PROGRESSION.  227 

From  what  has  just  been  said,  the  common  difference  of 
each  partial  progression  will  be  expressed  by 

h — a  c — h  d—e 


m+l  '       m+1   '       m+1 

which  are  equal  to  each  other,  since  a,  b,  c  .  .  .  are  in 
progression  :  therefore,  the  common  difference  is  the  same 
in  each  of  the  partial  progressions  ;  and  since  the  last  term 
of  the  first,  forms  the  Jlrst  term  of  the  second,  &c,  we  may 
conclude  that  all  of  these  partial  progressions  form  a  single 
progression. 


EXAMPLES. 

1 .  Find  the  sum  of  the  first  fifty  terms  of  the  progression 
2  .  9  .  16  .  23  .  .  . 

For  the  50th  term  we  have 

/z=2  +  49x7.~345. 


50 
Hence,        ^  =  (2  +  345)  x— =347x25  =  8675. 


2.  Find  the  100th  term  of  the  series  2  .  9  .  16  .  23  .  .  . 

Ans.  695. 

3.  P^ind  the  sum  of  100  terms  of  the  series  1.3.5. 
7.9  ...  Ans.   10000. 

4.  The  greatest  term  is  70,  the  common  difference  3,  and 
the  number  of  terms  21  :  what  is  the  least  term  and  the 
sum  of  the  series  ? 

Ans.  Least  term  10  ;  sum  of  series  840. 


228  FIRST  LESSONS  IN  ALGEBRA. 

5.  The  first  term  is  4,  the  common  difference  8,  and  the 

number  of  terms  8  :  what  is  the  last  term,  and  the  smn  of 

the  series  ? 

.  C  Last  term  60. 

Ans.     <  _ 

<  Sum     irz  256. 

6.  The  first  term  is  2,  the  last  term  20,  and  the  number 
of  terms  10  :  what  is  the  common  difference  ? 

A71S.  2. 

7.  Insert  four  means  between  the  two  numbers  4  and  19  : 
what  is  the  series  1 

Ans.    4  .  7  .   10  :   13  .   16  .   19. 


8.  The  first  term  of  a  decreasing  arithmetical  progres- 
sion is  10,  the  common  difference  — ,  and  the  number  of 

o 

terms  21  :  required  the  sum  of  the  series. 

Afis.   140. 

9.  In  a  progression  by  differences,  having  given  the 
common  difference  6,  the  last  term  185,  and  the  sum  of  the 
terms  2945  ;  find  the  first  term,  and  the  number  of  terms. 

Ans.  First  term  ==5  ;   number  of  terms  31. 

10.  Find  nine  arithmetical  means  between  each  antece- 
dent and  consequent  of  the  progression  2.5.8.11.14  ... 

Ans.  Common  dif,  or  r=:0,3. 

1 1 .  Find  the  number  of  men  contained  in  a  triangular  bat- 
talion, the  first  rank  containing  one  man,  the  second  2,  the 
third  3,  and  so  on  to  the  n^,  which  contains  n.  In  other 
words,  find  the  expression  for  the  sum  of  the  natural  num- 
bers 1,  2,  3  .  .  .,  from  I  to  n  inclusively. 

Ans.  «=^i). 


GEOMETRICAL    PROPORTION.  229 

9.  Find  the  sum  of  the  n  first  terms  of  the  progression 
of  uneven  numbers  1,  3,  5,  7,  9  .  .  .  Arts*  S  =  /i2. 

10.  One  hundred  stones  being  placed  on  the  ground  in  a 
straight  line,  at  the  distance  of  2  yards  from  each  other, 
how  far  will  a  person  travel  who  shall  bring  them  one  by 
one  to  a  basket,  placed  at  2  yards  from  the  first  stone  ? 

Ans.   11  miles,  840  yards. 


Geometrical  Proportion  and  Progression. 

144.  Ratio  is  the  quotient  arising  from  dividing  one 
quantity  by  another  quantity  of  the  same  kind.  Thus,  if 
the  numbers  3  and  6  have  the  same  unit,  the  ratio  of  3  to  6 
will  be  expressed  by 

And  in  general,  if  A  and  B  represent  quantities  of  the  same 
kind,  the  ratio  of  A  to  i?  will  be  expressed  by 

A' 

145.  If  there  be  four  numbers 

2,     4,     8,     16, 

having  such  values  that  the  second  divided  by  the  first  is 
equal  to  the  fourth  divided  by  the  third,  the  numbers  are 


QUEST.—144.  What  is  ratio  ]     What  is  the  ratio  of  3  to  6 1     Of  4 
to  121 

20 


230  FIRST  LESSONS  IN  ALGEBRA. 

said  to  be  in  proportion.  And  in  general,  if  there  be  four 
quantities,  A,  B,  C,  and  D,  having  such  values  that 

A-  C 

then  A  is  said  to  have  the  same  ratio  to  B  that  C  has  to  D ; 
or,  the  ratio  of  A  to  jB  is  equal  to  the  ratio  of  C  to  D, 
When  four  quantities  have  this  relation  to  each  other,  they 
are  said  to  be  in  proportion.  H-ence,  proportion  is  an  equality 
of  ratios. 

To  express  that  the  ratio  of  ^  to  5  is  equal  to  the  ratio 
of  C  to  Z),  we  write  the  quantities  thus ; 

A  :  B  :  :   C  :  D] 

and  read,  A  is  to  5  as  C  to  D. 

The  quantities  which  are  compared  together  are  called 
the  terms  of  the  proportion.  The  first  and  last  terms  are 
called  the  two  extremes,  and  the  second  and  third  terms,  the 
two  means.  Thus,  A  and  D  are  the  extremes,  and  B  and 
C  the  means. 

146.  Of  four  proportional  quantities,  the  first  and  third 
are  called  the  antecedents,  and  the  second  and  fourth  the 
consequents ;  and  the  last  is  said  to  be  a  fourth  proportional 
to  the  other  three  taken  in  order.  Thus,  in  the  last  pro- 
portion A  and  C  are  the  antecedents,  and  B  and  D  the 
consequents. 


Quest. — 145.  What  is  proportion  1  How  do  you  express  that  four 
numbers  are  in  proportion  1  What  are  the  numbers  called  1  What  are 
the  first  and  fourth  called'?  What  the  second  and  third] — 146.  In  four 
proportional  quantities,  what  are  the  first  and  third  called  ]  What  the 
second  and  fourth  1 


GEOMETRICAL    PROPORTION.  231 

1 47 .  Three  quantities  are  in  proportion  when  the  first 
has  the  same  ratio  to  the  second  that  the  second  has  to  the 
third ;  and  then  the  middle  term  is  said  to  be  a  mean  pro- 
portional between  the  other  two.     For  example, 

3  :  6  :  :  6  :   12; 

and  6  is  a  mean  proportional  between  3  and  12. 

148.  Quantities  are  said  to  be  in  proportion  by  muer- 
sion,  or  inversely,  when  the  consequents  are  made  the  ante- 
cedents and  the  antecedents  the  consequents. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  8  :   16, 
the  inverse  proportion  would  be 

6  :   3  :  :   16  :   8. 

1 49.  Quantities  are  said  to  be  in  proportion  by  alterna- 
tion, or  alternately,  when  antecedent  is  compared  with  ante- 
cedent and  consequent  with  consequent. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  8  :   16, 
the  alternate  proportion  would  be 

3  :   8  :  :   6  :   16. 


Quest. — 147.  When  are  three  quantities  proportionaU  What  is  the 
middle  one  called  1 — 148.  When  are  quantities  said  to  be  in  proportion 
by  inversion,  or  inversely  \ — 149.  When  are  quantities  in  proportion  by 
aHemation  1 


232  FIRST  LESSONS  IN  ALGEBRA. 

150.  Quantities  are  said  to  be  in  proportion  by  compo- 
sition, when  the  sum  of  the  antecedent  and  consequent  is 
compared  either  with  antecedent  or  consequent. 

Thus,,  if  we  have  the  proportion 

2  :   4  :  :   8  ;   16, 
the  proportion  by  composition  would  be 

2  +  4  :  4  :  :   8+16  :   16; 
that  is,  6:4::         24  :   16. 

151.  Quantities  are  said  to  be  in  proportion  by  division, 
when  the  difference  of  the  antecedent  and  consequent  is 
compared  either  with  antecedent  or  consequent. 

Thus,  if  we  have  the  proportion 

3  :  9  :  :   12  :  36, 
the  proportion  by  division  will  be 

9-3  :  9  :  :  36-12  :  36; 
that  is,  6  :  9  :  :  24  :  36. 

152.  Equi-multiples  of  two  or  more  quantities  are  the 
products  which  arise  from  multiplying  the  quantities  by  the 
same  number. 

Thus,  if  we  have  any  two  numbers,  as  6  and  5,  and  mul- 
tiply them  both  by  any  number,  as  9,  the  equi-multiples  will 
be  54  and  45  ;  for 

6x9=54,     and     5x9=45. 


Quest. — 150.  When  are  quantities  in  proportion  by  composition'? 
— 151.  When  are  quantities  in  proportion  by  division? — 153.  What 
are  equi-multiples  of  two  or  more  quantities  1 


GEOMETRICAL    PROPORTION.  233 

Also,  mxA  and  mx  B  are  equi-multiples  of  A  and  D,  the 
common  multiplier  being  m. 

153.  Two  quantities,  A  and  B,  are  said  to  be  recipro- 
colly  proportional,  or  inversely  proportional,  when  one  in- 
creases in  the  same  ratio  as  the  other  diminishes.  When 
this  relation  exists,  either  of  them  is  equal  to  a  constant 
quantity  divided  by  the  other. 

Thus,  if  we  had  any  two  numbers,  as  2  and  4,  so  related 
to  each  other  that  if  we  divided  one  by  any  number  we  must 
multiply  the  other  by  the  same  number,  one  would  increase 
just  as  fast  as  the  other  would  diminish,  and  their  product 
would  be  constant. 

154.  If  we  have  the  proportion 

A  :  B  :  :   C  :  D, 

we  have  —=z—,  (Art.  145); 

and  by  clearing  the  equation  of  fractions,  we  have 

BC^AD, 

That  is.  Of  four  proportional  quantities,  the  product  of  the 
two  extremes  is  equal  to  the  product  of  the  two  means. 

This  general  principle  is  apparent  in  the  proportion  be- 
tween the  numbers 

2  :   10  :  :   12  :   60, 
which  gives  2  x  60zizl0x  12  =  120. 


Quest. — 153.  When  are  two  quantities  said  to  be  reciprocally  pro- 
portional 1 — 154.  If  four  quantities  are  proportional,  what  is  the  product 
of  the  two  means  equal  to  1 

20* 


234  FIRST  LESSONS  IN  ALGEBRA. 

155.  If  four  quantities,  A,  B,   C,  D,  are  so  related  to 
each  other  that 

AxD=:BxC, 

we  shall  also  have  -—=:•—-; 

A      G 

and  hence,  A  :  B  :  :   C  :  D. 

That  is  ;  If  the  product  of  two  quantities  is  equal  to  the  prO' 
duct  of  two  other  quantities,  two  of  them  may  he  made  the 
extremes,  and  the  other  two  the  means  of  a  proportion. 
Thus,  if  we  have 

2x8r=4x4, 

we  also  have 

2  :  4  :  :  4  :  8. 

1  56.  If  we  have  three  proportional  quantities 
A  :  B  :  :  B  :   C, 

we  have  — rr=^t- ; 

A      B 

hence,  B^=:AC. 

That  is  :    The  square  of  the  middle  term  is  equal  to  the  product 
of  the  two  extremes. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  6  :   12, 
we  shall  also  have 

6x6=z62=:3xl2  =  36. 

Quest. — 155.  If  the  product  of  two  quantities  is  equal  to  the  product 
of  two  other  quantities,  may  the  four  be  placed  in  a  proportion  1  How  1 
— 156.  If  three  quantities  are  proportional,  what  is  the  product  of  the 
extremes  equal  to  1 


GEOMETRICAL    PROPORTION.  2B5 

157.  If  we  liave 

A  :  B  :  :   C  :  D,   and  consequently    -r-r=-T7, 

C 
multiply  both   members  of  the  last  equation   by  -^,  we 

then  obtain, 

A~B' 

and  hence,  A  :   C  :  :  B  :  D. 

That  is  :    If  four  quantities  are  proportional,  they  will  he  in 
proportion  by  alternation. 

Let  us  take,  as  an  example, 

10  :    15  :  :  20  :   30. 
We  shall  have,  by  alternating  the  terms, 

10  :  20  :  :   15  :   30. 

158.  If  we  have 

A  :  B  ::   C  :  D    and     A  :  B  :  :  E  :  F, 
we  shall  also  have 

B      D        :,    B      F 

B      F 
hence,  —^  =-^=   and   C  :  D  :  :  E  :  F. 

L>  XL 

That  is  :    If  there  are  two  sets  of  proportions  having  an 


Quest. — 157.  If  four  quantities  are  proportional,  will  they  be  in  pro- 
portion by  alternation  1 


236  FIRST  LESSONS  IN  ALGEBRA. 

antecedent  and  consequent  in  the  one  equal  to  an  antecedent 
and  consequent  of  the  other,  the  remaining  terms  will  he  pro- 
portional. 

If  we  have  the  two  proportions 

2  :  6  :  :  8  :  2*4     and     2   :  6  :  :   10  :  30, 

we  shall  also  have 

8  :  24  :  :   10  :  30. 

150.  If  we  have 

B     D 

A  :  B  :  :   C  :  D,    and  consequently    —=:—-, 

we  have,  by  dividing  1  by  each  member  of  the  equation, 

A        C 

-—=-—,    and  consequently   B  :  A  :  :  D  :   C. 

That  is  :  Four  proportional  quantities  will  be  in  proportion^ 
lohen  taken  inversely. 

To  give  an  example  in  numbers,  take  the  proportion 

7  :   14  :  :   8  :   16  ; 

then,  the  inverse  proportion  will  be 

14  :  7  :  :   16  :  8, 

in  which  the  ratio  is  one-half. 

160.  The  proportion 

A  :  B  ::   C  :  D     gives     AxB=BxC. 


Quest. — 158.  If  you  have  two  sets  of  proportions  having  an  ante- 
cedent and  consequent  in  each,  equal;  what  will  follow  1 — 159.  If  four 
quantities  are  in  proportion,  will  they  be  in  proportion  when  taken  in- 
versely 1 


GEOMETRICAL    PROPORTION.  237 

To  each  member  of  the  last  equation  add  BxD.     We 
shall  tlien  have 

{A+B)xD={C+D)xB; 

and  by  separating  the  factors,  we  obtain 

A+B  :  B  :  :   C+D  :  D. 

If,  instead  of  adding,  we  subtract  BxD  from  both  mem- 
bers, we  have 

{A-B)xD={C^D)xB; 
which  gives 

A-^B  :  B  :  :   C-D  :  D, 

That  is  :    If  four  quantities  are  proportional,  they  will  he  in 
proportion  hy  composition  or  division. 

Thus,  if  we  have  the  proportion 

9  :  27  :  :   16  :  48, 

we  shall  have,  by  composition, 

9+27  :  27  :  :   16  +  48  :  48; 

that  is,  36  :  27  :  :  64  :  48, 

in  which  the  ratio  is  three-fourths. 

The  proportion  gives  us,  by  division, 

27-9  :  27  :  :  48-16  :  48; 

that  is,  18  :  27  :  :   32  :   48, 

in  which  the  ratio  is  one  and  one -half. 


Quest. — 160.  If  four  quantities  are  in  proportion,  will  they  be  in  pro- 
portion by  composition  1  Will  they  be  in  proportion  by  division  1  What 
is  the  difference  between  composition  and  division  1 


288  FIRST  LESSONS  IN  ALGEBRA. 

^      1 6 1 .  If  we  have 

B__D_ 

and  multiply  the  numerator  and  denominator  of  the  first 
member  by  any  number  m,  we  obtain 

-=— -     and     mA  :  mB  :  :    C  :  D. 


mA 

That  is  :  Equal  multiples  of  two  quantities  have  the  same 
ratio  as  the  quantities  themselves. 

For  example,  if  we  have  the  proportion 

5  :   10  :  :   12  :  24, 

and  multiply  the  first  antecedent  and  consequent  by  6,  we 
have 

30  :  60  :  :   12  :  24, 

in  which  the  ratio  is  still  2. 

162.  The  proportions 

A   :  B  ::    C  :  D     and     A  :  B  \  :  E  :  F, 

give  AxD:=iBxC     and     AxF=iBxE\ 

adding  and  subtracting  these  equations,  we  obtain 

A(P±F):=zB{C±E),     or     A  .  B  :  :   C±E  :  D±.F. 

That  is  :  If  0  and  D,  the  antecedent  and  consequent,  he  aug- 
mented or  diminished  hy  quantities  E  and  F,  which  have  the 
same  ratio  as  C  to  D,  the  resulting  quantities  will  also  have 
the  same  ratio. 


Quest. — 161.  Have  equal  multiples  of  two  quantities  the  same  ratio 
as  the  quantities'! — 162.  Suppose  the  antecedent  and  consequent  be 
augmented  or  diminished  by  quantities  having  the  same  ratio  '^ 


GEOMETRICAL    PROPORTION.  239 

Let  us  take,  as  an  example,  the  proportion 

9  :   18  :  :  20  :  40, 

in  which  the  ratio  is  2. 

If  we  augment  the  antecedent  and  consequent  by  15  and 
30,  which  have  the  same  ratio,  we  shall  have 

9+15  :   18  +  30  :  :  20  :  40  ; 

that  is,  24  :  48  :  :  20  :  40, 

in  which  the  ratio  is  still  2. 

If  we  diminish  the  second  antecedent  and  consequent  by 
the  same  numbers,  we  have 

9  :   18  :  :  20-15  :  40-30; 

that  is,  9  :   18  :  :  5  :   10, 

in  which  the  ratio  is  still  2. 

163.  If  we  have  several  proportions 

A  :  B  :  :   C  :  D,     which  gives     AxD  =  BxC, 
A  :  B  :  :  E  :  F,         „         „         AxF=:BxE, 
A  :  B  :  :   G  :  H,         „         „         AxH=BxG. 
&c,  &:c, 
we  shall  have,  by  addition, 

A{D-]-F-]-H)=B{C-hE-hG); 
and  by  separating  the  factors, 

A  :  B  :  :   C-\-E-j-G  :  D+F+H. 

That  is  :  In  any  number  of  'proportions  having  the  same 
ratio^  any  antecedent  will  he  to  its  consequent,  as  the  sum  of 
the  antecedents  to  the  sum  of  the  consequents. 


240  FIRST    LESSONS    IN    ALGEBRA. 

Let  us  take,  for  example, 

2  :  4  :  :  6  :   12     and     1   :  2  :  :  3  :  6,     SlQ 
Then,  2:4::  6  +  3  :   12  +  6; 

that  is,  2  :  4  :  :  9  :   18, 

in  which  the  ratio  is  still  2. 

164.  If  we  have  four  proportional  quantities 

A  :  B  :  :   C  :  D,     we  have     —=-—  ; 

A.      G 

and  raising  both  members  to  any  power,  as  n,  we  have 

and  consequently 

A"  :  5"  :  :    C"  :  D\ 

That  is  :    If  four  quantities  are  proportional ,  any  like  powers 
or  roots  will  he  proportional. 
If  we  have,  for  example, 

2  :  4  :  :   3  :  6, 
we  shall  have         2^   :    42  :  :  3^  :  6^  ; 
that  is,  4   :   16   :  :    9  :  36, 

in  which  the  terms  are  proportional,  the  ratio  being  4. 

165.  Let  there  be  two  sets  of  proportions, 

A  :  B  :  :    C  :  D,     which  gives     -—=::-—-, 

A      O 

F     H 
E  :  F  :  :   G  :  H,         „         „         ~e~'g' 

Quest. — 163.  In  any  number  of  proportions  having  the  same  ratio, 
how  will  any  one  antecedent  be  to  its  consequent  1 — 164.  In  four  pro- 
portional quantities,  how  are  like  powers  or  roots  1 


GEOMETRICAL    PROGRESSION.  241 

Multiply  them  together,  member  by  member,  we  have 

-4^=-^    which  gives     AE  :  BF  i:   CG  :  BH. 

That  is  :    In  two  sets  of  proportional  quantities,  the  products 
of  the  corresponding  terms  will  he  proportional. 

Thus,  if  we  have  the  two  proportions 


8 

16  : 

:  10  : 

20 

^nd 

3  : 

4  : 

:  6  : 

8 

we  shall  have 

24  , 

64  : 

:  60  : 

160 

Geometrical  Progression, 

166.  We  have  thus  far  only  required  that  the  ratio  of 
the  first  term  to  the  second  should  be  the  same  as  that  of 
the  third  to  the  fourth. 

If  we  impose  the  farther  condition,  that  the  ratio  of  the 
second  term  to  the  third  shall  also  be  the  same  as  that  of  the 
first  to  the  second,  or  of  the  third  to  the  fourth,  we  shall  have 
a  series  of  numbers,  each  one  of  which,  divided  by  the 
preceding  one,  will  give  the  same  ratio.  Hence,  if  any 
term  be  multiplied  by  this  quotient,  the  product  will  be  the 
succeeding  term.  A  series  of  numbers  so  formed  is  called 
a  geometrical  progression.     Hence, 

A  Geometrical  Progression,  or  progression  hy  quotients,  is 
a  series  of  terms,  each  of  which  is  equal  to  the  product  of 


Quest. — 165.  In  two  sets  of  proportions,  how  are  the  products  of  the 
corresponding  terms  1 

21 


242  FIRST  LESSONS  IN  ALGBBRA. 

that  which  precedes  it  by  a  constant  number,  which  number 
is  called  the  ratio  of  the  progression.     Thus, 

1   :  3  :  9  :  27  :  81   :  243,  (fee, 

is  a  geometrical  progression,  which  is  written  by  merely 
placing  two  dots  between  each  two  of  the  terms.     Also, 

64  :  32  :  16  :  8  :  4  :  2  :   1 

is  a  geometrical  progression,  in  which  the  ratio  is  one-half. 

In  the  first  progression  each  term  is  contained  three  times 
in  the  one  that  follows,  and  hence  the  ratio  is  3.  In  the 
second,  each  term  is  contained  one-half  times  in  the  one 
which   follows,   and  hence  the  ratio  is  one-half. 

The  first  is  called  an  increasing  progression,  and  the 
second  a  decreasing  progression. 

Let  «,  h,  c,  d,  e,  f,  .  .  .  be  numbers  in  a  progression  by 
quotients  ;  they  are  written  thus  : 

a  X  h  I  c  I  d  ',  e  I  f  I  g  .  .  . 

and  it  is  enunciated  in  the  same  manner  as  a  progression  by 
differences.  It  is  necessary,  however,  to  make  the  distinc- 
tion, that  one  is  a  series  of  equal  differences,  and  the  other 
a  series  of  equal  quotients  or  ratios.  It  should  be  remarked 
that  each  term  is  at  the  same  time  an  antecedent  and  a  con- 
sequent, except  the  first,  which  is  only  an  antecedent,  and 
the  last,  which  is  only  a  consequent. 


Quest. — 166.  What  is  a  geometrical  progression'?  What  is  the  ratio 
of  the  progression  1  If  any  term  of  a  progression  be  multiphed  by  the 
ratio,  what  will  the  product  be  1  If  any  term  be  divided  by  the  ratio, 
what  will  the  quotient  be  1  How  is  a  progression  by  quotients  written  1 
Which  of  the  terms  is  only  an  antecedent  1  Which  only  a  consequent  ^ 
How  may  each  of  the  others  be  considered  1 


GEOMETRICAL    PROGRESSION.  24S 

167.  Let  q  denote  the  ratio  of  the  progression 

a  \  h  \  c  \  d  ,  .  .\ 

q  being   >1   when  the  progression  is  increasing,  and  5'<1 
when  it  is  decreasing.     Then,  since 


b  c  d  e  ^ 

T=^'    T=?'    T=^'    T=^'  '^^' 


we  have 


hz=^aq,     c=zhqz=zaq^,     d=icq=zaq^,     e=^dq^:zaq^, 
'       f—eq  —  aq^  .   .   .; 

that  is,  the  second  term  is  equal  to  aq,  the  third  to  aq^,  the 
fourth  to  acf',  the  fifth  to  aq^,  &c  ;  and  in  general,  any  term 
w,  that  is,  one  which  has  n—1  terms  before  it,  is  expressed 
by     fl5^"-i. 

Let  I  be  this  term ;  we  then  have  the  formula 

by  means  of  which  we  can  obtain  any  term  without  being 
obliged  to  find  all  the  terms  which  precede  it.  Hence,  to 
find  the  last  term  of  a  progression,  we  have  the  following 

RULii 

L  Raise  the  ratio  to  a  power  whose  exponent  is  one  less  titan 
the  number  of  terms. 

II.  Multiply  the  power  thus  found  by  the  first  term  :  the 
product  will  be  the  required  term. 


Quest. — 167.  By  what  letter  do  we  denote  the  ratio  of  the  progres- 
sion 1  In  an  increasing  progression  is  q  greater  or  less  than  11  In  a 
decreasing  progression  is  q  greater  or  less  than  11  If  a  is  the  first  term 
and  q  the  ratio,  what  is  the  second  term  equal  to  1  What  the  third  1 
What  the  fourth  1  What  is  the  last  term  equal  to  1  Give  the  rule  for 
finding  the  last  term. 


244  FIRST    LESSONS    IN    ALGEBRA. 

EXAMPLES. 

1 .  Find  the  5th  term  of  the  progression 

2  :  4  :  8  :   16  .  . 
in  which  the  first  term  is  2  and  the  common  ratio  2. 
5th  term=2  X2^=2  X  16  =  32     Atis, 

2.  Find  the  8th  term  of  the  progression 

2  :  6  :   18  :  -54   .  .  . 

8th  term=2  X  3*^=2  X  2187=4374     Ans. 

3.  Find  the  6th  term  of  the  progression 

2  :  8  :  32  :   128  .  .  . 
6th  term =2  x  4^=2  X  2048=4096    Aris. 

4.  Find  the  7th  term  of  the  progression 

3  :  9  :  27  :  81   .  .  . 

7th  term=3  x  3^=3  X  729=2187     Ans. 

5.  Find  the  6th  term  of  the  progression 

4  :  12  :  36  :   108  .  .  , 
6th  term =4  x  3^=4  X  243  =  972     A7is. 

6.  A  person  agreed  to  pay  his  servant  1  cent  for  the  first 
day,  two  for  the  second,  and  four  for  the  third,  doubling 
every  day  for  ten  days  :  how  much  did  he  receive  on  the 
tenth  day?  Ans.  $5,12 


GEOMETRICAL    PROGRESSION.  245 

1.  What  is  the  8th  term  of  the  progression 
9  :  36  :   144  :  576  .  .  . 
8th  term=r9x4^  =  9x  16384  =  147456     Ans. 
8.  Find  the  12th  term  of  the  progression 

64  :   16  :  4  :   1    :  4-  •  •  • 
4 

/  1  \ii        43         1  1 

12th  term=:64(  —  )    z=z — — = — Ans. 

\4J         411       48       65536 

1G8.  We  will  now  proceed  to  determine  the  sum  of  n 
terms  of  the  progression 

a  :  b  :  c  :  d  :  e  :  f  :  .  .  .  :  i  :  k  :  I; 

I  denoting  the  nth.  term. 

We  have  the  equations  (Art.  167), 

hzzzaq,     c=zhq,     d^=:cq,     e:=idq,  .   .   .  k^ziq,     lz=zkq\ 

and  by  adding  them  all  together,  member  to  member,  we 
deduce 

Sum  of  \st  members.  Sum  of  2nd  members. 

h+c+d+e+  .  .  .  +k+l=z{a+h  +  c+d-{-  .  .  .  -\-i+k)q\ 

in  which  we  see  that  the  first  member  wants  the  first  term 
a,  and  the  polynomial  within  the  parenthesis  in  the  second 
member  wants  the  last  term  I.  Hence,  if  we  call  the  sum 
of  the  terms  S,  we  have 

S— a=:(S  — Z)9'=S$'— Zg,     or     ^q—^^lq—a\ 

whence  S = — — — -. 

q-l 


246  FIRST  LESSONS  IN  ALGEBRA. 

Therefore,  to  obtain  the  sum  of  the  terms  of  a  geometrical 
progression,  we  have  the  following 


RULE. 

I.  Multiply  the  last  term  hy  the  ratio. 

II.  Subtract  the  first  term  from  the  product. 

III.  Divide  the  remainder  hy  the  ratio  diminished  hy  unity ^ 
and  the  quotient  will  he  the  sum  of  the  series. 

1 .  Find  the  sum  of  eight  terms  of  the  progression 

2  :  6  :   18  :  54  :   162  .  .  .  2x3'7rrr4374. 

q  —  l  2 

2.  Find  the  sum  of  the  progression 

2  :  4  :  8  :  16  :  32. 

q-l  1 

3.  Find  the  sum  of  ten  terms  of  the  progression 

2  :  6  :   18  :  54  :   162  .  .  .  2x39=.39366. 

Ans.  59048. 

4.  What  debt  may  be  discharged  in  a  year,  or  twelve 
months,  by  paying  $1  the  first  month,  $2  the  second  month, 


Quest. — 168.  Give  the  rule  for  finding  the  sum  of  the  series.     What 
is  the  first  step  1     What  the  second  1     What  the  third  1 


GEOMETRICAL   PROGHESSION.  247 


$4  the  third  month,  and  so  on,  each  succeeding  payment 
being  double  the  last ;  and  what  will  be  the  last  payment  1 

^^^5  Debt,       .     .     $4095. 
'  <  Last  payment,  $2048. 

5.  A  gentleman  married  his  daughter  on  New  Year's  day, 
and  gave  her  husband  1^.  towards  her  portion,  and  was  to 
double  it  on  the  first  day  of  every  month  during  the  year : 
what  was  her  portion  ?  Ans.  jE^204  1 5^. 

6.  A  man  bought  10  bushels  of  wheat  on  the  condition 
that  he  should  pay  1  cent  for  the  1st  bushel,  3  for  the  second, 
9  for  the  third,  and  so  on  to  the  last :  what  did  he  pay  for 
th^  last  bushel  and  for  the  ten  bushels  ? 

A   ^   ^  Last  bushel,  $196,83. 
^^'  i  Total  cost,     $295,24. 

7.  A  man  plants  4  bushels  of  barley,  which,  at  the  first 
harvest,  produced  32  bushels ;  these  he  also  plants,  which, 
in  like  manner,  produce  8  fold  ;  he  again  plants  all  his  crop, 
and  again  gets  8  fold,  and  so  on  for  16  years :  what  is  his 
last  crop,  and  what  the  sum  of  the  series  ? 

j^^    i  Last,  140737488355328SW. 
^**   i  Sum,  160842843834660. 

169.  When  the  progression  is  decreasing,  we  have 
5'<1   and  Z<a ;  the  above  formula 

q—1 

for  the  sum  is  then  written  under  the  form 

S=^, 

in  order  that  the  two  terms  of  the  fraction  may  be  positive. 

Quest. — 163.  What  is  the  formula  for  the  sum  of  the  series  of  a 
decreasing  progression  1 


248  FIRST    LESSONS    IN    ALGEBRA. 

1.  Find  the  sum  of  the  terms  o£  the  progression 

32  :   16  :  8  :  4  :  2. 

32-2X—      ^. 

l—q  1  1 

"2"  T 

2.  Find  the  sum  of  the  first  twelve  terms  of  the  progression 
1  ../1\^^  1 


64  :  16  :  4  :  1  :  —  :...  :  64^—')  , 


65536 


64---4;— xi  256--  ^ 


a— ?5'_    65536  4_     65536_     65535 


1— ^       _3_  3      '  196608 

T 

Remark. — 170.  We  perceive  that  the  principal  diffi- 
culty consists  in  obtaining  the  numerical  value  of  the  last 
term,  a  tedious  operation,  even  w^hen  the  number  of  terms 
is  not  very  great. 

3.  Find  the  sum  of  6  terms  of  the  progression 

512  :  128  :  32  .  .  . 

Ans.  682^. 

4.  Find  the  sum  of  seven  terms  of  the  progression 

2187  :  729  :  243  .  .  . 

Ans.  3279. 

5.  Find  the  sum  of  six  terms  of  the  progression 

972  :  324  :   108  .  .  . 

Ans.  1456. 

6.  Find  the  sum  of  8  terms  of  the  progression 

147456  :  36864  :  9216  .  .  . 

Ans.  196605. 


GEOMETRICAL    PROGRESSION.  249 

Of  Progressions  having  an  infinite  number  j/"  terms, 

111,  Let  there  be  the  decreasing  progression 
a  :  b  :  c  :  d  :  e  :  f  I  ,  ,  . 
containing  an  indefinite  number  of  terms.     In  the  formula 

substitute  for  I  its  value  aq''~^  (Art.  167),  and  we  have 

a—aq"" 
l-q 

which  represents  the  sum  of  n  terms  of  the  progression. 
This  may  be  put  under  the  form 


l-q      l-q 

Now,  since  the  progression  is  decreasing,  ^  is  a  proper 
fraction ;  and  q""  is  also  a  fraction,  which  diminishes  as  n 
increases.  Therefore,  the  greater  the  number  of  terms  we 
take,  the  more  will X  q"  diminish,  and  consequent- 
ly the  more  Avill  the  partial  sum  of  these  terms  approximate 

to  an  equality  with  the  first  part  of  S,  that  is,  to . 

Finally,  when  ?i  is  taken  greater  than  any  given  number,  or 
fi=infinity,  then x^'"     will  be  less  than  any  given 

number,  or  will  become  equal  to  0  ;  and  the  expression 


1-? 

will  represent  the  true  value  of  the  sum  of  all  the  terms  of 
the  series.     Whence  we  may  conclude,  that  the  expression 


250  FIRST  LESSONS  IN  ALGEBRA. 

for  the  sum  of  the  terms  of  a  decreasing  progression,  in  which 
the  number  of  terms  is  infinitey  is 


l-q 

That  is,  equal  to  the  first  term  divided  by  1  minus  the  ratio. 

This  is,  properly  speaking,  the  limit  to  which  the  partial 
sums  approach,  by  taking  a  greater  number  of  terms  in  the 
progression.      The    difference   between   these    sums   and 

can  become  as  small  as  we  please,  and  will  only 


\-q 

become  nothing  when  the  number  of  terms  taken  is  infinite. 

EXAMPLES. 

1.  Find  the  sum  of 

1111  .  ,  . 

^   ^T  ^"9-^27  ^81     '^'^^^^'y- 

We  have  for  the  expression  of  the  sum  of  the  terms 

r.        a  13 


1-^      1_J:.       2' 
3 


Ans. 


The  error  committed  by  taking  this  expression  for  the 
value  of  the  sum  of  the  n  first  terms,  is  expressed  by 


First  take  w=5  ;  it  becomes 
_3 
"2 


it  becomes 
\  3  /       2.3*       162 


Quest. — 165.  When  the  progression  is  decreasing  and  the  number  of 
terms  infinite,  what  is  the  value  of  the  sum  of  the  series  \ 


GEOMETRICAL    PROGRESSION.  251 

"When  n=z6,  we  find 

3  /  1  \6_    1         1         1 
YKs)  ~  162  ^Y"  486  ' 

3 

Whence  we  see  that  the  error  committed,  when     ---     is 

taken  for  the  sum  of  a  certain  number  of  terms,  is  less  in 
proportion  as  this  number  is  greater. 

2.  Again  take  the  progression 

1   :  —  :  —  :  —  :  —  :  —  :  &c,  .  .  . 
2        4        8       16      32 

We  have  S=— ^=: —=2.     Ans. 

3.  What  is  the  sum  of  the  progression 

^'TO'     Too-'     TOOO'     loko-'  ^"'   to  infinity. 
S— L.-_L_-il   Ans 
10 

172.  In  the  several  questions  of  geometrical  progres- 
sion there  are  ^ve  numbers  to  be  considered : 

1st.  The  first  term, a. 

2nd.  The  ratio, q, 

3rd.  The  number  of  terms, n. 

4th.  The  last  term, I. 

5th.  The  sum  of  the  terms, S. 


Quest. — 166.  How  many  numbers  are  considered  in  geometrical  pro- 

gression  1     What  are  they  1 


252  FIRST    LESSONS    IN    ALGEBRA. 

173.  We  shall  terminate  this  subject  by  the  question, 

To  find  a  mean  proportional  between  any  two  numbers, 
as  m  and  n. 

Denote  the  required  mean  by  x.     We  shall  then  have 
(Art.  156), 

s^z=z     rnXn, 

and  hence  x  =  ^mxn. 

That  is,  Multiply  the  two  numbers  together^  and  extract  the 
square  root  of  the  product. 

1.  What  is  the  geometrical  mean  between  the  numbers 
2  and  8  ? 

Mean  —  ^8x2  =  V 16 = 4     Ans. 

2.  What  is  the  mean  between  4  and  16  ?  Ans.  8. 

3.  What  is  the  mean  between  3  and  27  ?  Ans.  9. 

4.  What  is  the  mean  between  2  and  72  ?  Ans.  12. 

5.  What  is  the  mean  between  4  and  64  ?  Ans.  16. 


Quest. — 167.  How  do  you  find  a  mean  proportional  between  two 
numbers 'I 


THE    END. 


UNiVEki 


«%^ 


111869 


■SB^' 


